Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 11 Adsorption and Colloids Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 11 Adsorption and Colloids

1. Choose the correct option.

Question A.
The size of colloidal particles lies between
a. 10-10 m and 10-9 m
b. 10-9 m and 10-6 m
c. 10-6 m and 10-4 m
d. 10-5 m and 10-2 m
Answer:
b. 10-9 m and 10-6 m

Question B.
Gum in water is an example of
a. true solution
b. suspension
c. lyophilic sol
d. lyophobic sol
Answer:
c. lyophilic sol

Question C.
In Haber process of production of ammonia K2O is used as
a. catalyst
b. inhibitor
c. promotor
d. adsorbate
Answer:
c. promotor

Question D.
Fruit Jam is an example of-
a. sol
b. gel
c. emulsion
d. true solution
Answer:
b. gel

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

2. Answer in one sentence :

Question A.
Name type of adsorption in which van der Waals focres are present.
Answer:
Physical adsorption or physisorption.

Question B.
Name type of adsorption in which compound is formed.
Answer:
Chemical adsoiption or chemisorption.

Question C.
Write an equation for Freundlich adsorption isotherm.
Answer:
Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P1/n (n > 1) ……(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

3. Answer the following questions:

Question A.
Define the terms:
a. Inhibition
b. Electrophoresis
c. Catalysis.
Answer:
a. Inhibition:
The phenomenon in which the rate of chemical reaction is reduced by an inhibitor is called inhibition.

b. Electrophoresis:
The movement of colloidal particles under an applied electric potential is called electrophoresis.

c. Catalysis:
The phenomenon of increasing the rate of a chemical reaction with the help of a catalyst is known as catalysis.

Question B.
Define adsorption. Why students can read blackboard written by chalks?
Answer:

  • Adsorption is the phenomenon of accumulation of higher concentration of ‘one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.
  • When we write on blackboard using chalk, the chalk particles get adsorbed on the surface of the blackboard.

Hence, students can read blackboard written by chalks.

Question C.
Write characteristics of adsorption.
Answer:
Following are the characteristics of adsorption:

  • Adsorption is a surface phenomenon.
  • It depends upon the surface area of the adsorbent.
  • It involves physical forces (van der Waals forces) or chemical forces (chemical or covalent bonds).
  • Adsorbate is always present in higher concentration on the surface of an adsorbent than in the bulk.
  • Adsorption is dependent on temperature (of the surface) and pressure (of adsorbate gas).
  • It takes place with the evolution of heat (with some exceptions).

Question D.
Distinguish between Lyophobic and Lyophilic sols.
Answer:
Lyophobic sols (colloids):

  1. Lyophobic sols are formed only by special methods.
  2. They are irreversible.
  3. These are unstable and hence, require traces of stabilizers.
  4. Addition of small amount of electrolytes causes precipitation or coagulation of lyophobic sols.
  5. Viscosity of lyophobic sol is nearly the same as the dispersion medium.
  6. Surface tension of lyophobic sol is nearly the same as the dispersion medium.

Lyophilic sols (colloids):

  1. Lyophilic sols are formed easily by direct mixing.
  2. They are reversible.
  3. These are self-stabilized.
  4. Addition of large amount of electrolytes causes precipitation or coagulation of lyophilic sols.
  5. Viscosity of lyophilic sol is much higher than that of the dispersion medium.
  6. Surface tension of lyophilic sol is lower than that of dispersion medium.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

Question E.
Identify dispersed phase and dispersion medium in the following colloidal dispersions.
a. milk
b. blood
c. printing ink
d. fog
Answer:

Colloidal dispersion Dispersed phase Dispersion medium
Milk Liquid Liquid
Blood Solid Liquid
Printing ink Solid Liquid
Fog Liquid Gas

Question F.
Write notes on :
a. Tyndall effect
b. Brownian motion
c. Types of emulsion
d. Hardy-Schulze rule
Answer:
a. Tyndall effect:
i. Tyndall observed that when light passes through true solution, the path of light through it cannot be detected.
ii. However, if the light passes through a colloidal dispersion, the particles scatter some light in all directions and the path of the light through colloidal dispersion becomes visible to observer standing at right angles to its path.
iii. The phenomenon of scattering of light by colloidal particles and making path of light visible through the dispersion is referred as Tyndall effect and the bright cone of the light is called Tyndall cone.
iv. Tyndall effect is observed only when the following conditions are satisfied.

  • The diameter of the dispersed particles is not much smaller than the wavelength of light used.
  • The refractive indices of dispersed phase and dispersion medium differ largely.

v. Significance of Tyndall effect:

  • It is useful in determining number of particles in colloidal system and their particle size.
  • It is used to distinguish between colloidal dispersion and true solution.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 1

b. Brownian motion:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

  • Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
  • Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
  • This kinetic energy brings about Brownian motion.

c. Types of emulsion:
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

d. Hardy-Schulze rule:
i. Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.
ii. In the coagulation of negative sol, the flocculating power follows the following order:
Al3+ > Ba2+ > Na+
iii. Similarly, in the coagulation of positive sol, the flocculating power is in the following order:
[Fe (CN)6]4- > PO43- > SO42- > Cl

Question G.
Explain Electrophoresis in brief with the help of diagram. What are its applications ?
Answer:
i. Electrophoresis: Electrophoresis set up is shown in the diagram below.
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 2

  • The diagram shows U tube set up in which two platinum electrodes are dipped in a colloidal solution.
  • When electric potential is applied across two electrodes, colloidal particles move towards one or other electrode.
  • The movement of colloidal particles under an applied electric potential is called electrophoresis.
  • Positively charged particles move towards cathode while negatively charged particles migrate towards anode and get deposited on the respective electrode.

ii. Applications of electrophoresis:

  • On the basis of direction of movement of the colloidal particles under the influence of electric field, it is possible to know the sign of charge on the particles.
  • It is also used to measure the rate of migration of sol particles.
  • Mixture of colloidal particles can be separated by electrophoresis, since different colloidal particles in mixture migrate with different rates.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

Question H.
Explain why finely divided substance is more effective as adsorbent?
Answer:

  • Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon surface area of the adsorbent.
  • Adsorption increases with increase in surface area of the adsorbent.
  • Finely divided powdered substances provide larger surface area for a given mass.

Hence, finely divided substance is more effective as adsorbent.

Question I.
What is the adsorption Isotherm?
Answer:
The relationship between the amount of a substance adsorbed per unit mass of adsorbent and the equilibrium pressure (in case of gas) or concentration (in case of solution) at a given constant temperature is called an adsorption isotherm.

Question J.
Aqueous solution of raw sugar, when passed over beds of animal charcoal, becomes colourless. Explain.
Answer:

  • When aqueous solution of raw sugar is passed over beds of animal charcoal, charcoal adsorbs the coloured particles from the raw sugar.
  • Thus, due to the adsorption of coloured particles, raw sugar becomes colourless when passed over beds of animal charcoal.

Question K.
What happens when a beam of light is passed through a colloidal sol?
Answer:
i. When a beam of light is passed through colloidal sol, it is observed that the colloidal particles scatter some of the incident light in all directions.
ii. Because of this scattering of light, the path of light through the colloidal dispersion becomes visible to observer standing at right angles to its path and the phenomenon is known as Tyndall effect.
iii.
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 3

Question L.
Mention factors affecting adsorption of gas on solids.
Answer:
Adsorption of gases on solids depends upon the following factors:

  • Nature of adsorbate (gas)
  • Nature of solid adsorbent
  • Surface area of adsorbent
  • Temperature of the surface
  • Pressure of the gas

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

Question M.
Give four uses of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

  • The solid catalysts are used in many industrial manufacturing processes.
  • For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H2SO4 (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

  • It is a device which consists of activated charcoal or mixture of adsorbents.
  • It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.
iv. Production of high vacuum:

  • Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
  • A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question N.
Explain Bredig’s arc method.
Answer:

  • Colloidal sols can be prepared by electrical disintegration using Bredig’s arc method.
  • This process involves vaporization as well as condensation.
  • Colloidal sols of metals such as gold, silver, platinum can be prepared by this method.
  • In this method, electric arc is struck between electrodes of metal immersed in the dispersion medium.
  • The intense heat produced vapourizes the metal which then condenses to form particles of colloidal sol.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 4

Question O.
Explain the term emulsions and types of emulsions.
Answer:
i. A colloidal system in which one liquid is dispersed in another immiscible liquid is called an emulsion.
ii. There are liquid-liquid colloidal systems in which both liquids are either completely or partially immiscible.
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

4. Explain the following :

Question A.
A finely divided substance is more effective as adsorbent.
Answer:

  • Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon the surface area of the adsorbent.
  • Adsorption increases with an increase in surface area of the adsorbent.
  • Finely divided powdered substances provide a larger surface area for a given mass. Hence, a finely divided substance is more effective as an adsorbent.

Question B.
Freundlich adsorption isotherm, with the help of a graph.
Answer:
Graphical representation of the Freundlich adsorption isotherm:
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 5
i. Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P1/n (n > 1) ………(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.
ii. The graphical representation of Freundlich equation is as shown in the adjacent plot of x/m vs ‘P’.
iii. In case of solution, P in the equation (i) is replaced by the concentration (C) and thus,
\(\frac{x}{\mathrm{~m}}\) = k C1/n ………(ii)
iv. By taking logarithm on both sides of the equation (ii),
we get
log \(\frac{x}{\mathrm{~m}}\) = log k + \(\frac{1}{n}\) log C ……..(iii)
v. On plotting a graph of log \(\frac{x}{\mathrm{~m}}\) against log C or log P, a straight line is obtained as shown in the adjacent plot. The slope of the straight line is and intercept on Y-axis is log k.
vi. The factor \(\frac{1}{n}\) ranges from 0 to 1. Equation (iii) holds good over limited range of pressures.
a. When \(\frac{1}{n}\) → 0, \(\frac{x}{\mathrm{~m}}\) → constant, the adsorption is independent of pressure.
b. When \(\frac{1}{n}\) = 1, \(\frac{x}{\mathrm{~m}}\) = k P, i.e., \(\frac{x}{\mathrm{~m}}\) ∝ P, the adsorption varies directly with pressure.
c. The experimental isotherms tend to saturate at high pressure.
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 13

5. Distinguish between the following :

Question A.
Adsorption and absorption. Give one example.
Answer:
Adsorption:

  • Adsorption is a surface phenomenon as adsorbed matter is concentrated only at the surface and does not penetrate through the surface to the bulk of adsorbent.
  • Concentration of the adsorbate is high only at the surface of the adsorbent.
  • It is dependent on temperature and pressure.
  • It is accompanied by evolution of heat known as heat of adsorption.
  • It depends on surface area.
    e.g. Adsoiption of a gas or liquid like acetic acid by activated charcoal.

Absorption:

  • Absorption is a bulk phenomenon as absorbed matter is uniformly distributed inside as well as at the surface of the bulk of substance.
  • Concentration of the absorbate is uniform throughout the bulk of the absorbent.
  • It is independent of temperature and pressure.
  • It may or may not be accompanied by any evolution or absorption of heat.
  • It is independent of surface area.
    e.g. Absorption of water by cotton, absorption of ink by blotting paper.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

Question B.
Physisorption and chemisorption. Give one example.
Answer:
Physisorption:

  1. In physisorption, the forces operating are weak van der Waals forces.
  2. It is not specific in nature as all gases adsorb on all solids. For example, all gases adsorb on charcoal.
  3. The heat of adsorption is low and lies in the range 20-40 kJ mol-1.
  4. It occurs at low temperature and decreases with an increase of temperature.
  5. It is reversible.
  6. Physisorbed layer may be multimolecular layer of adsorbed particles under high pressure.
    e.g. At low temperature N2 gas is physically adsorbed on iron.

Chemisorption:

  1. In chemisorption, the forces operating are of chemical nature (covalent or ionic bonds).
  2. It is highly specific and occurs only when chemical bond formation is possible between adsorbent and adsorbate. For example, adsorption of oxygen on tungsten, hydrogen on nickel, etc.
  3. The heat of adsorption is high and lies in the range 40-200 kJ mol-1.
  4. It is favoured at high temperature, however, the extent of chemical adsorption is lowered at very high temperature due to bond breaking.
  5. It is irreversible.
  6. Chemisorption forms monomolecular layer of adsorbed particles.
    e.g. N2 gas chemically adsorbed on iron at high temperature forms a layer of iron nitride, which desorbs at very high temperature.

6. Adsorption is surface phenomenon. Explain.
Answer:
Consider a surface of a liquid or a solid.

  • The molecular forces at the surface of a liquid are unbalanced or in unsaturation state.
  • In solids, the ions or molecules at the surface of a crystal do not have their forces satisfied by the close contact with other particles.
  • Because of the unsaturation, solid and liquid surfaces tend to attract gases or dissolved substances with which they come in close contact. Thus, the substance accumulates on the surface of solid or liquid i.e., the substance gets adsorbed on the surface.

Hence, adsorption is a surface phenomenon.
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 6

7. Explain how the adsorption of gas on solid varies with
a. nature of adsorbate and adsorbent
b. surface area of adsorbent
Answer:
i. a. Nature of adsorbate:
1. All solids adsorb gases to some extent. It is observed that gases having high critical temperature liquify easily and can be readily adsorbed.
2. The gases such as SO2, Cl2, NH3 which are easily liquefiable are adsorbed to a larger extent as compared to gases such as N2, O2, H2, etc. which are difficult to liquify.
3. Thus, the amount of gas adsorbed by a solid depends on the nature of the adsorbate gas i.e., whether it is easily liquefiable or not.

b. Nature of adsorbent: Substances which provide large surface area for a given mass are effective as adsorbents and adsorb appreciable volumes of gases.
e.g. Silica gel and charcoal are effective adsorbents due to their porous nature.

ii. Surface area of the adsorbent:

  • Adsorption is a surface phenomenon. Hence, the extent of adsorption increases with increase in surface area of the adsorbent.
  • Finely divided substances, rough surfaces, colloidal substances are good adsorbents as they provide larger surface area for a given mass.

Note: Critical temperature of some gases and volume adsorbed.
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 7

8. Explain two applications of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

  • The solid catalysts are used in many industrial manufacturing processes.
  • For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H2SO4 (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

  • It is a device which consists of activated charcoal or mixture of adsorbents.
  • It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

9. Explain micelle formation in soap solution.
Answer:

  • Soap molecule has a long hydrophobic hydrocarbon chain called tail which is attached to hydrophilic ionic carboxylate group, called head.
  • In water, the soap molecules arrange themselves to form spherical particles that are called micelles.
  • In each micelle, the hydrophobic tails of soap molecules point to the centre and the hydrophilic heads lie on the surface of the sphere.
  • As a result of this, soap dispersion in water is stable.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 8

10. Draw labelled diagrams of the following :
a. Tyndall effect
b. Dialysis
c. Bredig’s arc method
d. Soap micelle
Answer:
a. Tyndall effect:
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 9

b. Dialysis:
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 10

c. Bredig’s arc method:
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 11

d. Soap micelle:
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 12

Activity :
Collect the information about methods to study surface chemistry.
Answer:
Following are the few methods that are employed to study surface chemistry.
i. X-ray photoelectron spectroscopy:
It is a surface-sensitive spectroscopic technique which is used to measure elemental composition of the surface, to determine elements that are present as contaminants on the surface, etc.

ii. Auger electron spectroscopy:
It is a common analytical technique which is used to study surfaces of materials.

iii. Temperature programmed desorption (TPD):
Adsorbed molecules get desorbed when the surface temperature is increased. TPD technique is used to observe these desorbed molecules and helps in providing information about binding energy between the adsorbate and adsorbent.

iv. Scanning Electron Microscopy:
In this technique, a scanning electron microscope is used to focus electron beam over the surface of the sample to be examined. The electron beam interacts with the sample and an image is obtained. This image provides information about surface structure and composition of the sample.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

11th Chemistry Digest Chapter 11 Adsorption and Colloids Intext Questions and Answers

Can you tell? (Textbook Page No. 160)

Question 1.
What is adsorption?
Answer:
Adsorption is the phenomenon of accumulation of higher concentration of one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.

Try this. (Textbook Page No. 161)

Question 1.
Dip a chalk in ink. What do you observe?
Answer:
When a chalk is dipped in ink, it is observed that the ink molecules are adsorbed at the surface of chalk and the surface becomes coloured, while the solvent of the ink goes deeper into the chalk due to absorption.

Internet my friend. (Textbook Page No. 172)

Question i.
Brownian motion
Answer:
Students can search relevant videos on YouTube to visualize Brownian motion.

Question ii.
Collect information about Brownian motion.
Answer:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

  • Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
  • Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
  • This kinetic energy brings about Brownian motion.

Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids

Internet my friend. (Textbook Page No. 172)

Question 1.
Collect information about surface chemistry.
Answer:

  • Surface or interface represents the boundary which separates two bulk phases.
    e.g. Boundary between water and its vapour is a liquid-gas interface.
  • Certain properties of substances, particularly of solids and liquids, depend upon the nature of the surface.
  • An interface usually has a thickness of a few molecules. However, its area depends on the size of the bulk phase particles.
  • Commonly considered bulk phases may be pure compounds or solutions.
  • A number of important phenomena, namely, dissolution, crystallization, heterogeneous catalysis, electrode processes and corrosion take place at an interface.
  • Thus, study of chemistry of surfaces is critical to many applications in industry, analytical investigations and day-to-day activities such as cleaning and softening of water.
  • The branch of chemistry which deals with the nature of surfaces and changes occurring on the surfaces is called surface chemistry.
  • Study of surfaces requires a rigorously clean surface. An ultra-clean metal surface can be obtained under very high vacuum, of the order of 10-8 to 10-9 pascal.
  • Adsorption, catalysis and colloids (such as emulsions and gels) are some of the important aspects of surface chemistry.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

Activity. (Textbook Page No. 172)

Question 1.
Calculate surface area to volume ratio of spherical particle. See how the ratio increases with the reduction of radius of the particle. Plot the ratio against the radius.
Answer:
The graph below shows that as the radius of the spherical particle decreases, the surface-to-volume ratio increases steadily.
Maharashtra Board Class 11 Chemistry Solutions Chapter 11 Adsorption and Colloids 14

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 10 States of Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 10 States of Matter

1. Select and write the most appropriate alternatives from the given choices.

Question A.
The unit of viscosity is
a. dynes
b. newton
c. gram
d. poise
Answer:
d. poise

Question B.
Which of the following is true for 2 moles of an ideal gas?
a. PV = nRT
b. PV = RT
c. PV = 2RT
d. PV = T
Answer:
c. PV = 2RT

Question C.
Intermolecular forces in liquid are
a. greater than gases
b. less than solids
c. both a and b
d. greater than solids
Answer:
c. both a and b

Question D.
Interactive forces are ………. in ideal gas.
a. nil
b. small
c. large
d. same as that of real gases
Answer:
a. nil

Question E.
At constant temperature the pressure of 22.4 dm3 volume of an ideal gas was increased from 105 kPa to 210 kPa, New volume could be-
a. 44.8 dm3
b. 11.2 dm3
c. 22.4 dm3
d. 5.6 dm3
Answer:
b. 11.2 dm3

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

2. Answer in one sentence.

Question A.
Name the term used for mixing of different gases by random molecular motion and ferquent collision.
Answer:
The mixing of different gases by random molecular motion and frequent collision is called diffusion.

Question B.
The pressure that each individual gas would exert if it were alone in the container, what do we call it as ?
Answer:
The pressure that each individual gas would exert if it were alone in the container is called as partial pressure.

Question C.
When a gas is heated the particles move more quickly. What is the change in volume of a heated gas if the pressure is kept constant ?
Answer:
The volume of the gas increases on heating if pressure is kept constant.

Question D.
A bubble of methane gas rises from the bottom of the North sea. What will happen to the size of the bubble as it rises to the surface ?
Answer:
According to Boyle’s law, the size of the bubble of methane gas increases as it rises to the surface.

Question E.
Convert the following temperatures from degree celcius to kelvin.
a. -15° C
b. 25° C
c. -197° C
d. 273° C
Answer:
a. T(K) = t°C +273.15
∴ T(K) = -15 °C + 273.15 = 258.15 K
b. T(K) = t°C +273.15
∴ T(K) = 25 °C + 273.15 = 298.15 K
c. T(K) = t°C + 273.15
∴ T(K) = -197 °C + 273.15 = 76.15 K
d. T(K) = t°C + 273.15
∴ T(K) = 273 °C + 273.15 = 546.15 K

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question F.
Convert the following pressure values into Pascals.
a. 10 atmosphere
b. 1 kPa.
c. 107000 Nm-2
d. 1 atmosphere
Answer:
a. 10 atmosphere:
1 atm = 101325 Pa
∴ 10 atm = 1013250 Pa
= 1.01325 × 106 Pa

b. 1 kPa:
1 kPa = 1000 Pa

c. 107000 N m-2:
1 N m-2 = 1 Pa
∴ 107000 Nm-2 = 107000 Pa
= 1.07 × 105 Pa

d. 1 atmosphere:
1 atm = 101325 Pa
= 1.01325 × 105 Pa

Question G.
Convert:
a. Exactly 1.5 atm to pascals
b. 89 kPa to newton per square metre (N m-2)
c. 101.325 kPa to bar
d. -100 °C to Kelvin
e. 0.124 torr to standard atmosphere
Answer:
a. Exactly 1.5 atm to pascals:
1 atm = 101325 Pa
∴ 1.5 atm = 1.5 × 101325
= 151987.5 Pa

b. 89 kPa to newton per square metre (N m-2):
1 Pa = 1 N m-2 and 1 Pa = 10-3 kPa
∴ 10-3 kPa = 1 N m-2
∴ 89 kPa = \(\frac{1 \times 89}{10^{-3}}\) N m-2 = 89000 N m-2

c. 101.325 kPa to bar:
1 bar = 1.0 × 105 Pa
= 1.0 × 102 k Pa
∴ 100 kPa = 1 bar
∴ 101.325 kPa = \(\frac{1 \times 101.325}{100}\)
= 1.01325 bar

d. -100 °C to Kelvin:
T(K) = t °C + 273.15
∴ T(K) = (- 100 °C) + 273.15 = 173.15 K

e. 0.124 torr to standard atmosphere:
1 atm = 760 torr
∴ 1 torr = \(\frac {1}{760}\)atm
∴ 0.124 torr = 0.124 × \(\frac {1}{760}\)
= 1.632 × 10-4 atm

Question H.
If density of a gas is measured at constant temperature and pressure then which of the following statement is correct ?
a. Density is directly proportional to molar mass of the gas.
b. Greater the density greater is the molar mass of the gas.
c. If density, temperature and pressure is given ideal gas equation can be used to find molar mass.
d. All the above statements are correct.
Answer:
d. All the above statements are correct.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question I.
Observe the following conversions.
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 1
Which of the above reactions is in accordance with the priciple of stoichiometry ?
Answer:
Both the reactions are in accordance with the principle of stoichiometry.
In the first reaction, both the reactants are completely consumed to form product according to reaction stoichiometry.
1 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride
In the second reaction, chlorine is the limiting reagent and it is completely consumed to form hydrogen chloride. Excess hydrogen remains unreacted at the end of the reaction. This reaction also follows principle of stoichiometry.
2 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride + 1 mol hydrogen

Question J.
Hot air balloons float in air because of the low density of the air inside the balloon. Explain this with the help of an appropriate gas law.
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 2
Answer:
The working of hot air balloon can be explained with the help of Charles’ law. According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the air inside the balloon expands and occupies more volume. Thus, hot air inside the balloon is less dense than the surrounding cold air. This causes the hot air balloon to float in air.

3. Answer the following questions.

Question A.
Identify the gas laws from the following diagrams.
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 3
Answer:
a. Boyle’s law
b. Charles’ law
c. Avogadro’s law [Note: Assuming, T constant]

Question B.
Consider a sample of a gas in a cylinder with a movable piston.
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 4
Show digramatically the changes in the position of piston, if
a. Pressure is increased from 1.0 bar to 2.0 bar at constant temperature.
b. Temperature is decreased from 300 K to 150 K at constant pressure
c. Temperature is decreased from 400 K to 300 K and pressure is decreased from 4 bar to 3 bar.
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 5
Thus, the volume of the gas remains the same.
Hence, there will be no change in the position of the piston.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question D.
List the characteristic physical properties of the gases.
Answer:
Characteristic physical properties of the gases:

  • Gases are lighter than solids and liquids (i.e., possess lower density).
  • Gases do not possess a fixed volume and shape. They occupy entire space available and take the shape of the container.
  • Gas molecules are widely separated and are in continuous, random motion. Therefore, gases exert pressure equally in all directions due to collision of gas molecules, on the walls of the container.
  • In case of gases, intermolecular forces are weakest.
  • Gases possess the property of diffusion, which is a spontaneous homogeneous inter mixing of two or more gases.
  • Gases are highly compressible.

Question E.
Define the terms:
a. Polarizability
b. Hydrogen bond
c. Aqueous tension
d. Dipole moment
Answer:
a. Polarizability is defined as the ability of an atom or a molecule to form momentary dipoles, that means, the ability of the atom or molecule to become polar by redistributing its electrons.

b. The electrostatic force of attraction between positively polarised hydrogen atom of one molecule and a highly electronegative atom (which may be negatively charged) of other molecule is called as hydrogen bond.

c. The pressure exerted by saturated water vapour is called aqueous tension.

d. Dipole moment (p) is the product of the magnitude of the charge (Q) and the distance between the centres of positive and negative charge (r). It is designated by a Greek Letter (p) and its unit is Debye (D).

Question F.
Would it be easier to drink water with a straw on the top of the Mount Everest or at the base ? Explain.
Answer:
When you drink through a straw, the pressure inside the straw reduces (as the air is withdraw by mouth) and the liquid is pushed up to your mouth by atmospheric pressure. Thus, drinking with a straw makes use of pressure difference to force the liquid into your mouth. So, if the pressure difference is less it will be difficult to drink through a straw. On the top of the Mount Everest, atmospheric pressure is very low. Hence, it will be difficult to drink water with a straw on the top of Mount Everest as compared to at the base.

Question G.
Identify type of the intermolecular forces in the following compounds.
a. CH3 – OH
b. CH2 = CH2
c. CHCl3
d. CH2Cl2
Answer:
a. Hydrogen bonding (dipole-dipole attraction) and London dispersion forces
b. London dispersion forces
c. Dipole-dipole interactions and London dispersion forces
d. Dipole-dipole interactions and London dispersion forces

Question H.
Name the types of intermolecular forces present in Ar, Cl2, CCl4 and HNO3.
Answer:
a. Ar: London dispersion forces
b. Cl2: London dispersion forces
c. CCl4: London dispersion forces
d. HNO3: Flydrogen bonding (dipole-dipole attraction) and London dispersion forces.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question I.
Match the pairs of the following :

A B
a. Boyle’s law i. At constant pressure and volume
b. Charles’ law ii. At constant temperature
iii. At constant pressure

Answer:
a – ii,
b – iii

Question J.
Write the statement for :
(a) Boyle’s law
(b) Charles’ law
Answer:
a. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

b. Statement for Charles’ law:
‘At constant pressure, the volume of a fixed mass of a gas is directly proportional to its temperature in Kelvin.

Question K.
Differentiate between Real gas and Ideal gas.
Answer:
Ideal gas:

  1. Strictly obeys Boyle’s and Charles’ law.
    \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) = 1
  2. Molecules are perfectly elastic.
  3. No attraction or repulsion between the gas molecules i.e. collision without loss of kinetic energy (K.E.)
  4. Actual volume of the gas molecules is negligible as compared to total volume of the gas.
  5. Ideal gases cannot be liquified even at low temperature but continues to obey Charles’ law and finally occupies zero volume at 0 K.
  6. Practically, ideal gas does not exist.

Real gas:

  1. Shows deviation from Boyle’s and Charles’ law at high pressure and temperature, i.e. obeys Boyle’s law and Charles’ law at low pressure and high temperature. \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) ≠ 1
  2. Molecules are not perfectly elastic.
  3. Intermolecular attraction is present, hence collision takes place with loss of kinetic energy.
  4. Actual volume of individual gas molecule is significant at high pressure and low- temperature.
  5. Real gases undergo liquefaction at low’ temperature when cooled and compressed.
  6. Gases that exist in nature like H2, O2, CO2, N2, He, etc. are real gases.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

4. Answer the following questions

Question A.
State and write mathematical expression for Dalton’s law of partial pressure and explain it with suitable example.
Answer:
i. Statement: The total pressure of a mixture of two or more non-reactive gases is the sum of the partial pressures of the individual gases in the mixture.
ii. Explanation:
Dalton’s law can be mathematically expressed as:
PTotal = P1 + P2 + P3 …(at constant T and V)
where, PTotal is the total pressure of the mixture and P1, P2, P3, … are the partial pressures of individual gases 1, 2, 3, … in the mixture.
For example, consider two non-reactive gases A and B. On mixing the two gases, pressure exerted by individual gas A in the mixture of both the gases is called partial pressure of gas A (say P1). Likewise, partial pressure of gas B is P2. According to Dalton’s law, total pressure of the mixture of gas A and B at constant T and V will be given as:
PTotal = P1 + P2

iii. Schematic illustration of Dalton’s law of partial pressures:
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 6

Question B.
Derive an Ideal gas equation. Mention the terms involved in it. Also write how it is utilised to obtain combined gas law.
Answer:
According to Boyle’s law,
V ∝ \(\frac{1}{\mathrm{P}}\) (at constant T and n) ……….(1)
According to Charles’ law,
V ∝ T (at constant P and n) ……(2)
According to Avogadro’s law,
V ∝ n (at constant P and T) ……(3)
Combining relations (1), (2) and (3), we get
V ∝ \(\frac{\mathrm{nT}}{\mathrm{P}}\)
Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get
∴ V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\)
On rearranging the above equation, we get
PV = nRT
where,
P = Pressure of gas,
V = Volume of gas,
n = number of moles of gas,
R = Gas constant,
T = Absolute temperature of gas.
This is the ideal gas equation or equation of state.
[Note: In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

The ideal gas equation is written as PV = nRT …(1)
On rearranging equation (1), we get,
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 7
The ideal gas equation used in this form is called combined gas law.

Question C.
With the help of graph answer the following –
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 8
At constant temperature,
a. Graph shows relation between pressure and volume. Represent the relation mathematically.
b. Identify the law.
c. Write the statement of law.
Answer:
a. P ∝ \(\frac{1}{\mathrm{~V}}\)
b. The graph represents Boyle’s law as it gives relation between pressure and volume at constant temperature.
c. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question D.
Write Postulates of kinetic theory of gases.
Answer:
Postulates of kinetic theory of gases:

  • Gases consist of tiny particles (molecules or atoms).
  • On an average, gas molecules remain far apart from each other. Therefore, the actual volume of the gas molecules is negligible as compared to the volume of the container. Hence, gases are highly compressible.
  • The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, gas expands to occupy entire volume of the container.
  • Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container.
  • Pressure of the gas is due to the collision of gas molecules with the walls of the container.
  • The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after collision.
  • The different gas molecules move with different velocities at any instant and hence have different kinetic energies. However, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

Question E.
Write a short note on
a. Vapour pressure.
b. Surface tension
c. Viscosity.
Answer:
a. Vapour pressure:

  • Molecules of liquid have tendency to escape from its surface to form vapour above it. This called evaporation.
  • When a liquid is placed in a closed container, the liquid undergoes evaporation and vapours formed undergo condensation.
  • At equilibrium, the rate of evaporation and rate of condensation are equal.
  • The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure or simply vapour pressure.
  • Vapour pressure is measured by means of a manometer.
  • The most common unit for vapour pressure is torr. 1 torr = 1 mm Hg.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 9
[Note: i. The vapour pressure of water is also called aqueous tension.
ii. Water has a vapour pressure of approximately 20 torr at room temperature.]

b. Surface tension:

  • The particles in the bulk of liquid are uniformly attracted in all directions and the net force acting on the molecules present inside the bulk is zero.
  • But the molecules at the surface experience a net attractive force towards the interior of the liquid, or the forces acting on the molecules on the surface are imbalanced.
  • Therefore, liquids have tendency to minimize their surface area and the surface acts as a stretched membrane.
  • The force acting per unit length perpendicular to the line drawn on the surface of liquid is called surface tension.
  • Unit: Surface tension is measured in SI unit, N m-1 and is denoted by Greek letter ‘γ’

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 10

c. Viscosity:
i. Liquids (fluids) have tendency to flow.
ii. Viscosity measures the magnitude of internal friction in a liquid or fluid to flow as measured by the force per unit area resisting uniform flow.
iii. Different layers of a liquid flow with different velocity. This called laminar flow. Here, the layers of molecules in the immediate contact of the fixed surface remains stationary. The subsequent layers slip over one another. Strong intermolecular forces obstruct the layers from slipping over one another, resulting in a friction between the layers.
iv. Viscosity is defined as the force of friction between the successive layers of a flowing liquid. It is also the resistance to the flow of a liquid.
v. When a liquid flow through a tube, the central layer has the highest velocity, whereas the layer along the inner wall in the tube remains stationary. This is a result of the viscosity of a liquid. Hence, a velocity gradient exists across the cross-section of the tube.
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 11
vi. Viscosity is expressed in terms of coefficient of viscosity, ‘η’ (Eta). The SI unit of viscosity coefficient is N s m-2 (newton second per square meter). In CGS system, the unit (η) is measured in poise.
1 poise = 1 g cm-1 s-1 = 10-1 kg m-1 s-1

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

5. Solve the following

Question A.
A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?
Answer:
Given: P1 = Initial pressure = 1 bar
V1 = Initial volume = 2.27 L
P2 = Final pressure = 0.3 bar
To find: V2 = Final volume
Formula: P1V1 = P2V2 (at constant n and T)
Calculation: According to Boyle’s law,
P1V1 = P2V2 (at constant n and T)
∴ V2 = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 2.27}{0.3}\) = 7.566667 L ≈ 7.567 L
Ans: The balloon can stay inflated below the volume of 7.567 L.

Question B.
A syringe has a volume of 10.0 cm3 at pressure 1 atm. If you plug the end so that no gas can escape and push the plunger down, what must be the final volume to change the pressure to 3.5 atm?
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 12
Answer:
Given: P1 = Initial pressure = 1 atm
V1 = Initial volume = 10.0 cm3
P2 = Final pressure = 3.5 atm
To find: V2 = Final volume
Formula: P1V1 = P2V2 (at constant n and T)
Calculation: According to Boyle’s law,
P1V1 = P2V2 (at constant n and T)
∴ V2 = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2}}=\frac{1 \times 10.0}{3.5}\)
= 2.857 L
Ans: The final volume of the gas in the syringe is 2.857 L.

Question C.
The volume of a given mass of a gas at 0°C is 2 dm3. Calculate the new volume of the gas at constant pressure when
a. The temperature is increased by 10°C.
b.The temperature is decreased by 10°C.
Answer:
Given: T1 = Initial temperature = 0 °C = 0 + 273.15 = 273.15 K,
V1 = Initial volume = 2 dm3
a. T2 = Final temperature = 273.15 K + 10 = 283.15 K
b. T2 = Final temperature = 273.15 K – 10 = 263.15 K
To find: V2 = Final volume in both the cases
Formula: \(\frac{\mathrm{V}_{\mathrm{l}}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 13
Ans: The new volume of a given mass of gas is:
a. 2.073 dm3
b. 1.927 dm3

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question D.
A hot air balloon has a volume of 2800 m3 at 99 °C. What is the volume if the air cools to 80 °C?
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 14
Answer:
Given: V1 = Initial volume = 2800 m3, T1 = Initial temperature = 99 °C = 99 + 273.15 = 372.15 K,
T2 = Final temperature = 80 °C = 80 + 273.15 K = 353.15 K
To find: V2 = Final volume
Formula: = \(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
∴ \(\mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~T}_{2}}{\mathrm{~T}_{1}}=\frac{2800 \times 353.15}{372.15}=\mathbf{2 6 5 7 \mathrm { m } ^ { 3 }}\)
Ans: The volume of the balloon when the air cools to 80 °C is 2657 m3.

Question E.
At 0 °C, a gas occupies 22.4 liters. How nuch hot must be the gas in celsius and in kelvin to reach volume of 25.0 literes?
Answer:
V1 = Initial volume of the gas = 22.4 L,
T1 = Initial temperature = 0 + 273.15 = 273.15 K,
V2 = Final volume = 25.0 L
To find: T2 = Final temperature in Celsius and in Kelvin
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 15
Ans: The temperature of the gas must be 31.7 °C or 304.9 K.

Question F.
A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177 °C and 1.25 mol of additional He gas was added to it?
Answer:
Given: V1 = Initial volume = 20 L, n1 = Initial number of moles = 0.650 mol
P1 = Initial pressure = 628.3 bar
T1 = Initial temperature = 37 °C = 37 + 273.15 K = 310.15 K
n2 = Final number of moles = 0.650 + 1.25 = 1.90 mol, V2 = Final volume = 12 L
T2 = Final temperature = 177 °C = 177 + 273.15 K = 450.15 K, R = 0.0821 L atm K-1 mol-1
To find: P2 = Final pressure
Formula: PV = nRT
Calculation: According to ideal gas equation,
P2V2 = n2RT2.
∴ \(\mathrm{P}_{2}=\frac{\mathrm{n}_{2} \mathrm{RT}_{2}}{\mathrm{~V}_{2}}=\frac{1.90 \times 0.0821 \times 450.15}{12}=\mathbf{5 . 8 5 2} \mathrm{atm}\)
Ans: The final pressure of the gas is 5.852 atm.
[Note: In the above numerical, converting the pressure value to different units, we get: 5.852 atm = 4447.52 torr = 5.928 bar]

Question G.
Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.
Answer:
Given: V = 2.32 L, P = 4.7 atm, T = 32 °C = 32 + 273.15 K = 305.15 K
R = 0.0821 L atm K-1 mol-1
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
∴ \(\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{4.7 \times 2.32}{0.0821 \times 305.15}=\mathbf{0 . 4 3 5} \mathrm{moles}\)
Ans: Number of moles of N2 gas in the given volume is 0.435 moles.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question H.
At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10 °C and volume of the gas 640 mL ?
Answer:
Given: V1 = Initial volume = 600 mL, V2 = Final volume = 640 mL
P1 = Initial pressure = 760 mm Hg
T1 = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
T2 = Final temperature = 10 °C = 10 + 273.15 K = 283.15 K
P2 = Final pressure
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law.
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 16
Ans: The final pressure of a gas is 676.654 mm Hg.

Question I.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of the gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture?
Answer:
Given: mO2 = 70.6 g, mNe = 167.5 g,
PTotal = 25 bar
To find: Partial pressure of each gas
Formula: P1 = x1 × PTotal
Calculation: Determine number of moles (n) of each gas using formula: n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 17
Ans: The partial pressure of dioxygen and neon are 5.2 bar and 19.8 bar respectively.

Question J.
Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm3 container at 20.0 °C.
Answer:
Given: n = number of moles = 1.0 mol, V = volume = 2.0 dm3
T = Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
R = 0.0821 L atm K-1 mol-1
To find: Pressure (P)
Formula: PV = nRT
Calculation: According to ideal gas equation,
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 18
Ans: The pressure of the given helium gas is 12.03 atm.

Question K.
Calculate the volume of 1 mole of a gas at exactly 20 °C at a pressure of 101.35 kPa.
Answer:
Given: n = number of moles = 1 mol, P = pressure = 101.35 kPa = 1.00025 atm ≈ 1 atm
T = Temperature = 20 °C = 20 + 273.15 K = 293.15 K
R = 0.0821 dm3 atm K-1 mol-1
To find: Volume (V)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter 19
Ans: The volume of the given gas is 24.07 dm3.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question L.
Calculate the number of molecules of methane in 0.50 m3 of the gas at a pressure of 2.0 × 102 kPa and a temperature of exactly 300 K.
Answer:
V = 0.5 m3, P = 2.0 × 102 kPa = 2.0 × 105 Pa
T = 300 K, R = 8.314 J K-1 mol-1
To find: Number of molecules of methane gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{2.0 \times 10^{5} \times 0.5}{8.314 \times 300}=40 \mathrm{~mol}\)
Number of molecules = n × NA = 40 × 6.022 × 1023 = 2.4088 × 1023 ≈ 2.409 × 1025
Ans: The number of molecules of methane gas present is 2.409 × 1025 molecules.

11th Chemistry Digest Chapter 10 States of Matter Intext Questions and Answers

Do you know? (Textbook Page No. 140)

Question 1.
Consider three compounds: H2S, H2Se and H2O. Identify which has the highest boiling point. Justify.
Answer:
Among the three compounds H2O, H2S and H2Se, the first one, H2O has the smallest molecular mass. But it has the highest B.P. of 100 °C. B.P. of H2S is -60 °C and of H2Se is -41.25 °C. The extraordinary high B.P. of H2O is due to very strong hydrogen bonding even though it has the lowest molecular mass.

Can you tell? (Textbook Page No. 140)

Question i.
What are the various components present in the atmosphere?
Answer:
Various components present in the atmosphere are as follows:
a. Nitrogen (78%)
b. Oxygen (21%)
c. Carbon dioxide and other gases (0.03%)
d. Inert gases (mainly argon) (0.97%)
e. Traces of water vapour

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Question ii.
Name five elements and five compounds those exist as gases at room temperature.
Answer:
Five elements and five compounds that exist as gases at room temperature are as follows:

No. Elements
a. Nitrogen
b. Oxygen
c. Hydrogen
d. Chlorine
e. Argon
No. Compounds
a. Carbon dioxide
b. Carbon monoxide
c. Nitrogen dioxide
d. Sulphur dioxide
e. Methane

Just think. (Textbook Page No. 140)

Question 1.
What is air?
Answer:

  • Air is a mixture of various gases.
  • One cannot see air but can feel the cool breeze.
  • The composition of air by volume is around 78 percent N2, 21 percent O2 and 1 percent other gases including CO2.

Use your brainpower. (Textbook Page No. 141)

Question 1.
Find the unit in which car-tyre pressure is measured.
Answer:
Car-tyre pressure is measured in the units of pounds per square inch (psi) or Newton per metre square (N m-2).

Do you know? (Textbook Page No. 142)

Question 1.
How does a bicycle pump work?
Answer:
A bicycle pump works on Boyle’s law. Pushing a bicycle pump squashes the same number of particles into a smaller volume. This squashing means particles hit the walls of the pump more often, increasing the pressure. The increased pressure of a gas can be felt on palm by pushing in the piston of a bicycle pump.

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Internet my friend (Textbook Page No. 143)

Question 1.
i. Watch Boyle’s law experiment.
ii. Find applications of Boyle’s law.
iii. Try to study how Boyle’s law helps in ‘scuba-diving’ i.e., importance of Boyle’s law in scuba diving an exhilarating sport.
Answer:
i. Students can refer to ‘Boyle’s law experiment’ on YouTube channel of ‘Socratica’.
ii. a. Syringes: When the plunger of a syringe is pulled back out, it causes the volume of the gas inside it to increase due to the reduction of pressure. This creates a vacuum in the syringe, which is constantly trying to adjust the pressure back to normal. However, since the only substance available, such as the blood or medication, is on the other side of the needle, this liquid is sucked into the vacuum, increasing the pressure and decreasing the volume of the gas. When we push the plunger back down, the pressure again increases, lowering the volume inside the syringe, and pushing the fluid out.

b. Respiration: Boyle’s law is essential for the human breathing process. When person breathes in, his/her lung volume increases and the pressure within decreases. Since air always moves from areas of high pressure to areas of low pressure, air is drawn into the lungs. The opposite happens when person exhales. Since the lung volume decreases, the pressure within increases, forcing the air out of the lungs

c. Storage of Gases: Many industries store gases under high pressure. This allows the gas to be stored at a low volume, saving plenty of storage space.
[Note: Students are expected to search more on the internet about various other applications of Boyle’s law on their own.]

iii. Importance of Boyle’s law in scuba diving:
a. Boyle’s law affects scuba diving in many ways.
b. It explains the role of pressure in the aquatic environment.
c. As divers descend, the water pressure surrounding them increases, causing air in their body and equipment to have a smaller volume. As the divers ascend, water pressure decreases, causing their body and equipment to expand to acquire a greater volume.
d. Furthermore, it is crucial that scuba divers never attempt to hold their breath when immersed in water.
e. According to Boyle’s law, if divers attempt this when they ascend to a body of water of less pressure, then the air that is trapped in their lungs will over-expand and rupture. This is known as Pulmonary Barotrauma. Thus, it is important for scuba divers to exhale as they ascend because the external pressure increases.
f. Also, if a diver returns to the surface too quickly, dissolved gases in the blood expand and form bubbles, which can get stuck in capillaries and organs (causing the ‘bends’).
[Note: Students are expected to collect additional information their own.]

Just think. (Textbook Page No. 144)

Question i.
Why does bicycle tyre burst during summer?
Answer:

  • According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the volume also increases.
  • During summer, the temperature of the surrounding air is high. Due to the high temperature, the air inside the tyre gets heated. This will increase the volume of the tyres and it will burst.

Question ii.
Why do the hot air balloons fly high?
Answer:

  • According to Charles’ law, at constant pressure, gases expand on heating and become less dense. Thus, hot air is less dense than cold air.
  • In a hot air balloon, the air inside it is heated by a burner. Upon heating, the air inside the balloon expands and becomes lighter (less dense) than the cooler air on the outside. This causes the hot air balloon to fly high in air.

Just think. (Textbook Page No. 145)

Question 1.
i. List out various real-life examples of Charles’ law.
ii. Refer and watch Charles’ law experiments.
Answer:
i. Few real-life examples of Charles’ law:
a. Helium balloon: If we fill a helium balloon in a warm or hot room, and then take it into a cold room, it shrinks up and will look like it has lost some of the air inside it. This shows that gases expand on heating and contract on cooling.
b. A bottle of deodorant: If we expose a bottle of deodorant to sunlight and high temperatures, the air molecules inside the bottle will expand which can lead to the bursting of the deodorant bottle. This is another example of Charles’ law.
c. Basketball: You may have noticed that a basketball has less responsive bounce during winter than in summer. This yet another example of Charles’ law. When a basketball is inflated, the air pressure inside it is set to a fixed value. As the temperature falls, the volume of the gas inside the ball also decreases proportionally.
[Note: Students are expected to collect additional real-life examples on their own,]

ii. pi [Note: Students can scan the adjacent QR code to visualize Charles’ law with the aid of a relevant video.]

Maharashtra Board Class 11 Chemistry Solutions Chapter 10 States of Matter

Use your brainpower. (Textbook Page No. 146)

Question 1.
Why does the pressure in the automobile tyres change during hot summer or winter season?
Answer:

  • According to Gay-Lussac’s law, at constant volume, pressure of a fixed amount of a gas is directly proportional to its absolute temperature.
  • During hot summer, the temperature of automobile tyre increases faster. Consequently, the air inside the tyre gets heated and the gas molecules starts moving faster.
  • As the volume of the tyre remains constant, the pressure inside it increases.
  • During winter, the temperature of automobile tyre decreases. Consequently, the air inside the tyre gets cooled and the gas molecules starts moving much slower and the pressure inside the tyre decreases.

Just think. (Textbook Page No. 149)

Question 1.
Do all pure gases and mixtures of gases obey the gas laws?
Answer:
Yes, the gas laws are also applicable to the mixtures of gases. The measurable properties of a mixture of the gases such as pressure, temperature, volume, and amount of gaseous mixture are all related by an ideal gas law.

Just think. (Textbook Page No. 150)

Question 1.
Where is Dalton’s law applicable?
Answer:
Air is a gaseous mixture of different gases. Dalton’s law is useful for the study of various phenomena in air, for example, air pollution.

Just think. (Textbook Page No. 155)

Question 1.
What makes the oil rise through the wick in an oil lamp?
Answer:
In an oil lamp, oil rises through the wick due to the capillary action. Such a capillary rise of oil is due to the surface tension of oil. The wick acts as a capillary tube. When the wick is placed in oil, the attractive forces between the oil and the inner wall of the capillary (wick) pull the oil up through the wick.

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 9 Elements of Group 13, 14 and 15 Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

1. Choose the correct option.

Question A.
Which of the following is not an allotrope of carbon?
a. buckyball
b. diamond
c. graphite
d. emerald
Answer:
d. emerald

Question B.
………… is inorganic graphite.
a. borax
b. diborane
c. boron nitride
d. colemanite
Answer:
c. boron nitride

Question C.
Haber’s process is used for the preparation of ………….
a. HNO3
b. NH3
c. NH2CONH2
d. NH4OH
Answer:
b. NH3

Question D.
Thallium shows a different oxidation state because ……………
a. of inert pair effect
b. it is an inner transition element
c. it is metal
d. of its high electronegativity
Answer:
a. of inert pair effect

Question E.
Which of the following shows the most prominent inert pair effect?
a. C
b. Si
c. Ge
d. Pb
Answer:
d. Pb

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

2. Identify the group 14 element that best fits each of the following description.

A. Non-metallic element
B. Form the most acidic oxide
C. They prefer +2 oxidation state.
D. Forms strong π bonds.
Answer:
i. Carbon (C)
ii. Carbon
iii. Tin (Sn) and lead (Pb)
iv. Carbon

3. Give reasons.

A. Ga3+ salts are better reducing agent while Tl3+ salts are better oxidising agent.
B. PbCl4 is less stable than PbCl2
Answer:
A. i. Both gallium (Ga) and thallium (Tl) belong to group 13.
ii. Ga is lighter element compared to thallium Tl. Therefore, its +3 oxidation state is stable. Thus, Ga+ loses two electrons and get oxidized to Ga3+. Hence, Ga+ salts are better reducing agent.
iii. Thallium is a heavy element. Therefore, due to the inert pair effect, Tl forms stable compounds in +1 oxidation state. Thus, Tl3+ salts get easily reduced to Tl1+ by accepting two electrons. Hence, Tl3+ salts are better oxidizing agent.
[Note: This question is modified so as to apply the appropriate textual concept.]

B. i. Pb has electronic configuration [Xe] 4f14 5d10 6s2 6p2.
ii. Due to poor shielding of 6s2 electrons by inner d and f electrons, it is difficult to remove 6s2 electrons (inert pair).
iii. Thus, due to inert pair effect, +2 oxidation state is more stable than +4 oxidation state.
Hence, PbCl4 is less stable than PbCl2.

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

4. Give the formula of a compound in which carbon exhibit an oxidation state of

A. +4
B. +2
C. -4
Answer:
A. CCl4
B. CO
C. CH4

5. Explain the trend of the following in group 13 elements :

A. atomic radii
B. ionization enthalpy
C. electron affinity
Answer:
A. Atomic radii:

  • In group 13, on moving down the group, the atomic radii increases from B to Al.
  • However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.
  • However, the atomic radii again increases from Ga to Tl.
  • Therefore, the atomic radii of the group 13 elements varies in the following order:
    B < Al > Ga < In < Tl

B. Ionization enthalpy:

  • Ionization enthalpies show irregular trend in the group 13 elements.
  • As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.
  • However, there is a marginal difference in the ionization enthalpy from Al to Tl.
  • The ionization enthalpy increases slightly for Ga but decreases from Ga to In.
    In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly.
  • Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.
  • The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.

C. Electron affinity:
a. Electron affinity shows irregular trend. It first increases from B to A1 and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.

b. From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.
Note: Electron affinity of group 13 elements:
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 1

6. Answer the following

Question A.
What is hybridization of Al in AlCl3?
Answer:
Al is sp2 hybridized in AlCl3.

Question B.
Name a molecule having banana bond.
Answer:
Diborane (B2H6)

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

7. Draw the structure of the following

Question A.
Orthophosphoric acid
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 2

Question B.
Resonance structure of nitric acid
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 3

8. Find out the difference between

Question A.
Diamond and Graphite
Answer:
Diamond:

  1. It has a three-dimensional network structure.
  2. In diamond, each carbon atom is sp3 hybridized.
  3. Each carbon atom in diamond is linked to four other carbon atoms.
  4. Diamond is poor conductor of electricity due to absence of free electrons.
  5. Diamond is the hardest known natural substance.

Graphite:

  1. It has a two-dimensional hexagonal layered structure.
  2. In graphite, each carbon atom is sp2 hybridized.
  3. Each carbon atom in graphite is linked to three other carbon atoms.
  4. Graphite is good conductor of electricity due to presence of free electrons in its structure.
  5. Graphite is soft and slippery.

Question B.
White phosphorus and Red phosphorus
Answer:
White phosphorus:

  1. It consists of discrete tetrahedral P4 molecules.
  2. It is less stable and more reactive.
  3. It exhibits chemiluminescence.
  4. It is poisonous.

Red phosphorus:

  1. It consists chains of P4 molecules linked together by covalent bonds.
  2. It is stable and less reactive.
  3. It does not exhibit chemiluminescence.
  4. It is nonpoisonous.

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

9. What are silicones? Where are they used?
Answer:
i. a. Silicones are organosilicon polymers having R2SiO (where, R = CH3 or C6H5 group) as a repeating unit held together by
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 4
b. Since the empirical formula R2SiO (where R = CH3 or C6H5 group) is similar to that of ketones (R2CO), these compounds are named as silicones.

ii. Applications: They are used as

  • insulating material for electrical appliances.
  • water proofing of fabrics.
  • sealant.
  • high temperature lubricants.
  • for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

10. Explain the trend in oxidation state of elements from nitrogen to bismuth.
Answer:

  • Group 15 elements have five valence electrons (ns2 np3). Common oxidation states are -3, +3 and +5. The range of oxidation state is from -3 to +5.
  • Group 15 elements exhibit positive oxidation states such as +3 and +5. Due to inert pair effect, the stability of +5 oxidation state decreases and +3 oxidation state increases on moving down the group.
  • Group 15 elements show tendency to donate electron pairs in -3 oxidation state. This tendency is maximum for nitrogen.
  • The group 15 elements achieve +5 oxidation state only through covalent bonding.
    e. g. NH3, PH3, ASH3, SbH3, and BiH3 contain 3 covalent bonds. PCl5 and PF5 contain 5 covalent bonds.

11. Give the test that is used to detect borate radical is qualitative analysis.
Answer:
i. Borax when heated with ethyl alcohol and concentrated H2SO4, produces volatile vapours of triethyl borate, which bum with green edged flame.
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 5
ii. The above reaction is Used as a test for the detection and removal of borate radical \(\left(\mathrm{BO}_{3}^{3-}\right)\) in qualitative analysis.

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

12. Explain structure and bonding of diborane.
Answer:

  • Electronic configuration of boron is 1s2 2s2 2p1. Thus, it has only three valence electrons.
  • In diborane, each boron atom is sp3 hybridized. Three of such hybrid orbitals are half filled while the fourth sp3 hybrid orbital remains vacant.
  • The two half-filled sp3 hybrid orbitals of each B atom overlap with 1s orbitals of two terminal H atoms and form four B – H covalent bonds. These bonds are also known as two-centred-two-electron (2c-2e) bonds.
  • When ‘1s’ orbital of each of the remaining two H atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three-centred-two- electron bonds (3c-2e) or banana bonds.
  • Hydrogen atoms involved in (3c-2e) bonds are the bridging H atoms i.e., H atoms in two B – H – B bonds.
  • In diborane, two B atoms and four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 6

13. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is white crystalline solid and used for inorganic qualitative analysis.

a. Name the compound produced.
b. Write the reaction that explains its formation.
Answer:
a. Borax
b. Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 7

14. Ammonia is a good complexing agent. Explain.
Answer:
i. The lone pair of electrons on nitrogen atom facilitates complexation of ammonia with transition metal ions. Thus, ammonia is a good complexing agent as it forms complex by donating its lone pair of electrons.
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 8
ii. This reaction is used for the detection of metal ions such as Cu2+ and Ag+.

15. State true or false. Correct the false statement.

A. The acidic nature of oxides of group 13 increases down the graph.
B. The tendency for catenation is much higher for C than for Si.
Answer:
A. False
The acidic nature of oxides of group 13 decreases down the group. It changes from acidic through amphoteric to basic.
B. True

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

16. Match the pairs from column A and B.

Column A Column B
i. BCl3 a. Angular molecule
ii. SiO2 b. Linear covalent molecule
iii. CO2 c. Tetrahedral molecule
d. Planar trigonal molecule

Answer:
i – d,
ii – c,
iii – b

17. Give the reactions supporting basic nature of ammonia.
Answer:
In the following reactions ammonia reacts with acids to form the corresponding ammonium salts which indicates basic nature of ammonia.
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 9

18. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
Answer:
i. When silver nitrate (AgNO3) is added to an aqueous solution of salt sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed.
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 10

ii. On adding ammonium hydroxide (NH4OH) to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 11

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

11th Chemistry Digest Chapter 9 Elements of Group 13, 14 and 15 Intext Questions and Answers

Can you recall? (Textbook Page No. 123)

Question 1.
If the valence shell electronic configuration of an element is 3s2 3p1, in which block of the periodic table is it placed?
Answer:
The element having valence shell electronic configuration 3s2 3p1 must be placed in the p-block of the periodic table as its last electron enters in p-subshell (3p).

Can you recall? (Textbook Page No. 127)

Question 1.
What is common between diamond and graphite?
Answer:
Both diamond and graphite are made up of carbon atoms as they are two allotropes of carbon.

Can you recall? (Textbook Page No. 129)

Question i.
Which element from the following pairs has higher ionization enthalpy?
B and TI, N and Bi
Answer:
Among B and Tl, boron has higher ionization enthalpy while, among N and Bi, nitrogen has higher ionization enthalpy.

Question ii.
Does boron form covalent compound or ionic?
Answer:
Yes, boron forms covalent compound.

Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

Try this. (Textbook Page No. 131)

Question 1.
Find out the structural formulae of various oxyacids of phosphorus.
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 12
Maharashtra Board Class 11 Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15, 13

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 8 Elements of Group 1 and 2 Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 8 Elements of Group 1 and 2

1. Explain the following

Question A.
Hydrogen shows similarity with alkali metals as well as halogens.
Answer:

  • The electronic configuration of hydrogen is 1s1 which is similar to the outer electronic configuration of alkali metals of group 1 i.e., ns1.
  • However, 1s1 also resembles the outer electronic configuration of group 17 elements i.e., ns2 np5.
  • By adding one electron to H, it will attain the electronic configuration of the inert gas He which is 1s2, and by adding one electron to ns2 np5 we get ns2 np6 which is the outer electronic configuration of the remaining inert gases.
  • Therefore, some chemical properties of hydrogen are similar to those of alkali metals while some resemble halogens.

Hence, hydrogen shows similarity with alkali metals as well as halogens.

Question B.
Standard reduction potential of alkali metals have high negative values.
Answer:

  • The general outer electronic configuration of alkali metals is ns1.
  • They readily lose one valence shell electron to achieve stable noble gas configuration and hence, they are highly electropositive and are good reducing agents.

Hence, standard reduction potentials of alkali metals have high negative values.

Question C.
Alkaline earth metals have low values of electronegativity; which decrease down the group.
Answer:

  • Electronegativity represents attractive force exerted by the nucleus on shared electrons.
  • The general outer electronic configuration of alkaline earth metals is ns2. They readily lose their two valence shell electrons to achieve stable noble gas configuration. They are electropositive and hence, they have low values of electronegativity.

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

Question D.
Sodium dissolves in liquid ammonia to form a solution which shows electrical conductivity.
Answer:
i. Sodium dissolves in liquid ammonia giving deep blue coloured solutions which is electrically conducting in nature.
Na + (x + y) NH3 → [Na(NH3)x]+ + [e(NH3)y]
ii. Due to formation of ions, the solution shows electrical conductivity.

Question E.
BeCl2 is covalent while MgCl2 is ionic.
Answer:

  • Be2+ ion has very small ionic size and therefore, it has very high charge density.
  • Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl) which is larger in size.
  • This results in partial covalent character of the bond in BeCl2.
  • Mg2+ ion has very less tendency to distort the electron cloud of Cl due to the bigger size of Mg2+ as compared to Be2+.

Hence, BeCl2 is covalent while MgCl2 is ionic.

Question F.
Lithium floats an water while sodium floats and catches fire when put in water.
Answer:

  • When lithium and sodium react with water, hydrogen gas is released. Due to these hydrogen gas bubbles, lithium and sodium floats on water.
    eg. 2Na + 2H2O → 2Na+ + 2OH + H2
  • The reactivity of group 1 metals increases with increasing atomic radius and lowering of ionization enthalpy down the group.
  • Thus, sodium having lower ionization enthalpy, is more reactive than lithium.
  • Hence, lithium reacts slowly while sodium reacts vigorously with water.
  • Since the reaction of sodium with water is highly exothermic, it catches fire when put in water.

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

2. Write balanced chemical equations for the following.

Question A.
CO2 is passed into concentrated solution of NaCl, which is saturated with NH3.
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 1

Question B.
A 50% solution of sulphuric acid is subjected to electrolyte oxidation and the product is hydrolysed.
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 2

Question C.
Magnesium is heated in air.
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 3

Question D.
Beryllium oxide is treated separately with aqueous HCl and aqueous NaOH solutions.
Answer:
Beryllium oxide (BeO) is an amphoteric oxide and thus, it reacts with both acid (HCl) as well as base (NaOH) to give the corresponding products.
i. \(\mathrm{BeO}+\underset{(\text { Acid })}{2 \mathrm{HCl}} \longrightarrow \mathrm{BeCl}_{2}+\mathrm{H}_{2} \mathrm{O}\)
ii. \(\mathrm{BeO}+\underset{(\text { Base })}{2 \mathrm{NaOH}} \longrightarrow \mathrm{Na}_{2} \mathrm{BeO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

3. Answer the following questions

Question A.
Describe the diagonal relationship between Li and Mg with the help of two illustrative properties.
Answer:
a. The relative placement of these elements with similar properties in the periodic table is across a diagonal and is called diagonal relationship.
b. Lithium is placed in the group 1 and period 2 of the modem periodic table. It resembles with magnesium which is placed in the group 2 and period 3.
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 4

ii. Li and Mg show similarities in many of their properties.
e. g.
a. Reaction with oxygen:
1. Group 1 elements except lithium, react with oxygen present in the air to form oxides (M2O) as well as peroxides (M2O2) and superoxides (MO2) on further reaction with excess of oxygen.
2. This anomalous behaviour of lithium is due to its resemblance with magnesium as a result of diagonal relationship.
3. As group 2 elements form monoxides i.e., oxides, lithium also form monoxides.
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 5

b. Reaction with nitrogen:
1. All the group 1 elements react only with oxygen present in the air to form oxides while group 2 elements react with both nitrogen and oxygen present in the air forming corresponding oxides and nitrides.
2. However, lithium reacts with oxygen as well as nitrogen present in the air due to its resemblance with magnesium.
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 6

Question B.
Describe the industrial production of dihydrogen from steam. Also write the chemical reaction involved.
Answer:
Three stages are involved in the industrial production of dihydrogen from steam.
i. Stage 1:
a. Reaction of steam on hydrocarbon or coke (C) at 1270 K temperature in presence of nickel catalyst gives water-gas which is a mixture of carbon monoxide and hydrogen.
1. Reaction of steam with hydrocarbon:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 7
2. Reaction of steam with coke or carbon (C):
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 8
b. Sawdust, scrapwood, etc. can also be used in place of carbon.

ii. Stage 2:
Water-gas shift reaction: When carbon monoxide in the water-gas reacts with steam in the presence of iron chromate (FeCrO4) as catalyst, it gets transformed into carbon dioxide. This is called water-gas shift reaction.
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 9

iii. Stage 3: In the last stage, carbon dioxide is removed by scrubbing with sodium arsenite solution.

Question C.
A water sample, which did not give lather with soap, was found to contain Ca(HCO3)2 and Mg(HCO3)2. Which chemical will make this water give lather with soap? Explain with the help of chemical reactions.
Answer:

  • Soap does not lather in hard water due to presence of the soluble salts of calcium and magnesium in it. So, the given water sample is hard water.
  • Hardness of hard water can be removed by removal of these calcium and magnesium salts.
  • Sodium carbonate is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates. Thus, it will make water give lather with soap.
    e.g. Ca(HCO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaHCO3(aq)

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

Question D.
Name the isotopes of hydrogen. Write their atomic composition schematically and explain which of these is radioactive ?
Answer:
i. Hydrogen has three isotopes i.e., hydrogen \(\left({ }_{1}^{1} \mathrm{H}\right)\), deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) and tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) with mass numbers 1, 2 and 3 respectively.
ii. They all contain one proton and one electron but different number of neutrons in the nucleus.
iii. Atomic composition of isotopes of hydrogen:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 10
iv. Tritium is a radioactive nuclide with half-life period 12.4 years and emits low energy β particles.
v. Schematic representation of isotopes of hydrogen is as follows:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 11

4. Name the following

Question A.
Alkali metal with smallest atom.
Answer:
Lithium (Li)

Question B.
The most abundant element in the universe.
Answer:
Hydrogen (H)

Question C.
Radioactive alkali metal.
Answer:
Francium (Fr)

Question D.
Ions having high concentration in cell sap.
Answer:
Potassium ions (K+)

Question E.
A compound having hydrogen, aluminium and lithium as its constituent elements.
Answer:
Lithium aluminium hydride (LiAlH4)

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

5. Choose the correct option.

Question A.
The unstable isotope of hydrogen is …..
a. H-1
b. H-2
c. H-3
d. H-4
Answer:
c. H-3

Question B.
Identify the odd one.
a. Rb
b. Ra
c. Sr
d. Be
Answer:
a. Rb

Question C.
Which of the following is Lewis acid ?
a. BaCl2
b. KCl
c. BeCl2
d. LiCl
Answer:
c. BeCl2

Question D.
What happens when crystalline Na2CO3 is heated ?
a. releases CO2
b. loses H2O
c. decomposes into NaHCO3
d. colour changes.
Answer:
b. loses H2O

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

Activity :

1. Collect the information of preparation of dihydrogen and make a chart.
2. Find out the s block elements compounds importance/uses.
Answer:
1.
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 12

2. Uses of s-block elements:
Group 1 elements (alkali metals):
a. Lithium: Lithium is widely used in batteries.
b. Sodium:

  • Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
  • Sodium is also used as an important reagent in the Wurtz reaction.
  • It is used in the manufacture of sodium vapour lamp.

c. Potassium:

  • Potassium has a vital role in biological system.
  • Potassium chloride (KCl) is used as a fertilizer.
  • Potassium hydroxide (KOH) is used in the manufacture of soft soaps and also as an excellent absorbent of carbon dioxide.
  • Potassium superoxide (KO2) is used as a source of oxygen.

d. Caesium: Caesium is used in devising photoelectric cells.

Group 2 elements (alkaline earth metals):
a. Magnesium: Magnesium hydroxide [Mg(OH)2] in its suspension form is used as an antacid.
b. Calcium: Compounds of calcium such as limestone and gypsum are used as constituents of cement and mortar.
c. Barium: BaSO4 being insoluble in H2O and opaque to X-rays is used as ‘barium meal’ to scan the X-ray of human digestive system.
[Note: Students are expected to collect additional information about preparation of dihydrogen and uses of s-block elements on their own.]

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

11th Chemistry Digest Chapter 8 Elements of Group 1 and 2 Intext Questions and Answers

Can you recall? (Textbook Page No. 110)

Question 1.
Which is the first element in the periodic table?
Answer:
Hydrogen is the first element in the periodic table.

Question 2.
What are isotopes?
Answer:
Many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element. Isotopes of an element have the same atomic number (number of protons) but different atomic mass numbers due to different number of neutrons in their nuclei.

Question 3.
Write the formulae of the compounds of hydrogen formed with sodium and chlorine.
Answer:
Hydrogen combines with sodium to form sodium hydride (NaH) while it reacts with chlorine to form hydrogen chloride (HCl).

Can you tell? (Textbook Page No. 110)

Question 1.
In which group should hydrogen be placed? In group 1 or group 17? Why?
Answer:

  • Hydrogen contains one valence electron in its valence shell and thus, its valency is one. Therefore, hydrogen resembles alkali metals (group 1 elements) as they also contain one electron in their valence shell (alkali metals tend to lose their valence electron).
  • However, hydrogen also shows similarity with halogens (group 17 elements) as their valency is also one because halogens tend to accept one electron in their valence shell.
  • Due to this unique behaviour, it is difficult to assign any definite position to hydrogen in the modem periodic table.

Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2

Just think! (Textbook Page No. 112)

Question 1.
\(2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaH}_{(\mathrm{s})}\)
In the above chemical reaction which element does undergo oxidation and which does undergo reduction?
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na(s) + H2(g) → 2Na+ + 2H
ii. Charge development suggests that each sodium atom loses one electron to form Na+ and each hydrogen atom gains one electron to form H. This can be represented as follows:
Maharashtra Board Class 11 Chemistry Solutions Chapter 8 Elements of Group 1 and 2, 13
iii. Na is oxidised to NaH by loss of electrons while the elemental hydrogen is reduced to NaH by a gain of electrons.

Can you recall? (Textbook Page No. 113)

Question i.
What is the name of the family of reactive metals having valency one?
Answer:
The family of reactive metals having valency one is known as alkali metals (group 1).

Question ii.
What is the name of the family of reactive metals having valency two?
Answer:
The family of reactive metals having valency two is known as alkaline earth metals (group 2).

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 7 Modern Periodic Table Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 7 Modern Periodic Table

1. Explain the following

Question A.
The elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.
Answer:

  • Li, B, Be and N belong to the same period.
  • As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily and therefore, electronegativity increases.

Hence, the elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.

Question B.
The atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.
Answer:

  • Cl, I and Br belong to group 17 (halogen group) in the periodic table.
  • As we move down the group from top to bottom in the periodic table, a new shell gets added in the atom of the elements.
  • As a result, the effective nuclear charge decreases due to increase in the atomic size as well as increased shielding effect.
  • Therefore, the valence electrons experience less attractive force from the nucleus and are held less tightly resulting in the increased atomic radius.
  • Thus, their atomic radii increases in the following order down the group.
    Cl (99 pm) < Br (114 pm) < I (133 pm)

Hence, the atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.

Question C.
The ionic radii of F and Na+ are 133 and 98 pm, respectively.
Answer:

  • F and Na+ are isoelectronic ions as they both have 10 electrons.
  • However, the nuclear charge on F is +9 while that of Na+ is +11.
  • In isoelectronic species, larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.

Thus, F has larger ionic radii (133 pm) than Na+ (98 pm).

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Question D.
13Al is a metal, 14Si is a metalloid and 15P is a nonmetal.
Answer:

  • Electronic configuration of Al is [Ne] 3s2 3p1, 14Si is [Ne] 3s2 3p2 and that of 15P is [Ne] 3s2 3p3.
  • Metals are characterized by the ability to form compounds by loss of valence electrons.
  • ‘Al’ has 3 valence electrons, thus shows tendency to lose 3 valence electrons to complete its octet. Hence, Al is a metal.
  • Nonmetals are characterized by the ability to form compounds by gain of valence electrons in valence shell.
  • ‘P’ has 5 valence electrons thus, shows tendency to gain 3 electrons to complete its octet. Hence, ‘P’ is a nonmetal.
  • Si has four valence electrons, thus it can either lose/gain electrons to complete its octet. Hence, behaves as a metalloid.

Question E.
Cu forms coloured salts while Zn forms colourless salts.
Answer:

  • Electronic configuration of 29CU is [Ar] 3d104s1 while that of Zn is [Ar] 3d104s2.
  • Electronic configuration of Cu in its +1 oxidation state is [Ar] 3d10 while that in +2 oxidation state is [Ar] 3d9.
  • Therefore, Cu contains partially filled d orbitals in +2 oxidation state and thus, Cu2+ salts are coloured.
  • However, Zn has completely filled d orbital which is highly stable and hence, it does not form coloured ions.

Hence, Cu forms coloured salts while Zn forms colourless salts.

2. Write the outer electronic configuration of the following using orbital notation method. Justify.
A. Ge (belongs to period 4 and group 14)
B. Po (belongs to period 6 and group 16)
C. Cu (belongs to period 4 and group 11)
Answer:
A. a. Ge belongs to period 4. Therefore, n = 4.
b. Group 14 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 14 elements is ns2 np2.
d. Thus, the outer electronic configuration of Ge is 4s2 4p2.

B. a. Po belongs to period 6. Therefore, n = 6.
b. Group 16 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 16 elements is ns2 np4.
d. Thus, the outer electronic configuration of Po is 6s2 6p4.

C. a. Cu belongs to period 4. Therefore, n = 4.
b. Group 11 indicates that the element belongs to the d-block of the modem periodic table.
c. The general outer electronic configuration of the d-block elements is ns0-2(n-1)d1-10.
d. The expected configuration of Cu is 4s23d9. However, the observed configuration of Cu is 4s13d10. This is due to the extra stability associated with completely filled d-subshell. Thus, the outer electronic configuration of Cu is 4s13d10.

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

3. Answer the following

Question A.
La belongs to group 3 while Hg belongs to group 12 and both belong to period 6 of the periodic table. Write down the general outer electronic configuration of the ten elements from La to Hg together using orbital notation method.
Answer:
i. La and Hg both belongs to period 6. Therefore, n = 6.
ii. Elements of group 3 to group 12 belong to the d-block of the modem periodic table.
iii. The general outer electronic configuration of the d-block elements is ns0-2 (n -1 )1-10.
iv. Therefore, the outer electronic configuration of all ten elements from La to Hg is as given in the table below.
Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table 1
[Note: There are 14 elements between La and Hf which are called lanthanides. Therefore, after La, electrons are filled in 4f subshell of lanthanide elements. Once all the 14 elements of lanthanide series are filled, next electron enters 5d subshell of Hf. Hence, the outer electronic configurations of Hf to Hg often include completely filled 4f subshell. For example, the electronic configuration of Hf ‘5d26s2’ can also be written as ‘4f145d26s2’.]

Question B.
Ionization enthalpy of Li is 520 kJ mol-1 while that of F is 1681 kJ mol-1. Explain.
Answer:

  • Both Li and F belong to period 2.
  • Across a period, the screening effect is the same while the effective nuclear charge increases.
  • As a result, the outer electron is held more tightly and therefore, the ionization enthalpy increases across a period.
  • Hence, F will have higher ionization enthalpy than Li.

Thus, ionization enthalpy of Li is 520 kJ mol-1 while that of F is 1681 kJ mol-1.

Question C.
Explain the screening effect with a suitable example.
Answer:
i. In a multi-electron atom, the electrons in the inner shells tend to prevent the attractive influence of the nucleus from reaching the outermost electron.
ii. Thus, they act as a screen or shield between the nuclear attraction and outermost or valence electrons. This effect of the inner electrons on the outer electrons is known as screening effect or shielding effect.
iii. Across a period, screening effect due to inner electrons remains the same as electrons are added to the same shell.
iv. Down the group, screening effect due to inner electrons increases as a new valence shell is added.
e.g. Potassium (19K) has electronic configuration 1s22s22p63s23p64s1.
K has 4 shells and thus, the valence shell electrons are effectively shielded by the electrons present in the inner three shells. As a result of this, valence shell electron (4s1) in K experiences much less effective nuclear charge and can be easily removed.

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Question D.
Why the second ionization enthalpy is greater than the first ionization enthalpy ?
Answer:
The second ionization enthalpy (ΔiH2) is greater than the first ionization enthalpy (ΔiH1) as it involves removal of electron from the positively charged species.

Question E.
Why the elements belonging to the same group do have similar chemical properties ?
Answer:

  • Chemical properties of elements depend upon their valency.
  • Elements belonging to the same group have the same valency.

Hence, the elements belonging to the same group show similar chemical properties.

Question F.
Explain : electronegativity and electron gain enthalpy. Which of the two can be measured experimentally?
Answer:
i. The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN). Electronegativity cannot be measured experimentally. However, various numerical scales to express electronegativity were developed by many scientists. Pauling scale of electronegativity is the one used most widely.

ii. Electron gain enthalpy is a quantitative measure of the ease with which an atom adds an electron forming the anion and is expressed in kJ mol-1. Thus, it is an experimentally measurable quantity.

4. Choose the correct option

Question A.
Consider the elements B, Al, Mg and K predict the correct order of metallic character :
a. B > Al > Mg > K
b. Al > Mg > B > K
c. Mg > Al > K > B
d. K > Mg > Al > B
Answer:
d. K > Mg > Al > B

Question B.
In modern periodic table, the period number indicates the :
a. atomic number
b. atomic mass
c. principal quantum number
d. azimuthal quantum number
Answer:
c. principal quantum number

Question C.
The lanthanides are placed in the periodic table at
a. left hand side
b. right hand side
c. middle
d. bottom
Answer:
d. bottom

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Question D.
If the valence shell electronic configuration is ns2np5, the element will belong to
a. alkali metals
b. halogens
c. alkaline earth metals
d. actinides
Answer:
b. halogens

Question E.
In which group of elements of the modern periodic table are halogen placed ?
a. 17
b. 6
c. 4
d. 2
Answer:
a. 17

Question F.
Which of the atomic number represent the s-block elements ?
a. 7, 15
b. 3, 12
c. 6, 14
d. 9, 17
Answer:
b. 3, 12

Question G.
Which of the following pairs is NOT isoelectronic ?
a. Na+ and Na
b. Mg2+ and Ne
c. Al3+ and B3+
d. P3 and N3-
Answer:
b. Mg2+ and Ne

Question H.
Which of the following pair of elements has similar properties ?
a. 13, 31
b. 11, 20
c. 12, 10
d. 21, 33
Answer:
a. 13, 31

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

5. Answer the following questions

Question A.
The electronic configuration of some elements are given below:
a. 1s2
b. 1s22s22p6
In which group and period of the periodic table they are placed ?
Answer:
a. 1s2
Here n = 1. Therefore, the element belongs to the 1st period.
The outer electronic configuration 1s2 corresponds to the maximum capacity of 1s, the complete duplet. Therefore, the element is placed at the end of the 1st period in the group 18 of inert gases in the modem periodic table,

b. 1s22s22p6
Here n = 2. Therefore, the element belongs to the 2nd period.
The outer electronic configuration 2s22p6 corresponds to complete octet. Therefore, the element is placed in the 2nd period of group 18 in the modem periodic table.

Question B.
For each of the following pairs, indicate which of the two species is of large size :
a. Fe2+ or Fe3+
b. Mg2+ or Ca2+
Answer:
a. Fe2+ has a larger size than Fe3+.
b. Ca2+ has a larger size than Mg2+.

Question C.
Select the smaller ion form each of the following pairs:
a. K+, Li+
b. N3-, F
Answer:
i. Li+ has smaller ionic radius than K+
ii. F has smaller ionic radius than N3-.

Question D.
With the help of diagram answer the questions given below:
Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table 2
a. Which atom should have smaller ionization enthalpy, oxygen or sulfur?
b. The lithium forms +1 ions while berylium forms +2 ions ?
Answer:
Sulfur should have smaller ionization energy than oxygen.
a. Lithium has electronic configuration 1s22s1 while that of beryllium is 1s22s2.
b. Li can achieve a noble gas configuration by losing one electron while Be can do so by losing two electrons. Hence, lithium forms +1 ions while beryllium forms +2 ions.

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Question E.
Define : a. Ionic radius
b. Electronegativity
Answer:
a. Ionic radius: Ionic radius is defined as the distance of valence shell of electrons from the centre of the nucleus in an ion.

b. Electronegativity: The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN).

Question F.
Compare chemical properties of metals and non-metals.
Answer:
i. Metals (like alkali metals) react vigorously with oxygen to form oxides which reacts with water to form strong bases.
e. g. Sodium (Na) reacts with oxygen to form Na2O which produces NaOH on reaction with water.

ii. Nonmetals (like halogens) react with oxygen to form oxides which on reaction with water form strong acids.
e.g. Chlorine reacts with oxygen to form Cl2O7 which produces HClO4 on reaction with water.

Question G.
What are the valence electrons ? For s-block and p-block elements show that number of valence electrons is equal to its group number.
Answer:

  • Electrons present in the outermost shell of the atom of an element are called valence electrons.
  • 3Li is an s-block element and its electronic configuration is 1s22s1. Since it has one valence electron, it is placed in group 1.
  • Therefore, for s-block elements, group number = number of valence electrons.
  • However, for p-block elements, group number = 18 – number of electrons required to attain complete octet.
  • 7N is a p-block element and its electronic configuration is 1s22s22p3. Since it has five electrons in its valence shell, it is short of three electrons to complete its octet.
  • Therefore, its group number = 18 – 3 = 15.

Question H.
Define ionization enthalpy. Name the factors on which ionisation enthalpy depends? How does it vary down the group and across a period?
Answer:
i. The energy required to remove an electron from the isolated gaseous atom in its ground state is called ionization enthalpy (ΔiH).
Ionization enthalpy is the quantitative measure of tendency of an element to lose electron and expressed in kJ mol-1.

ii. Ionization energy depends on the following factors

  • Size (radius) of an atom
  • Nuclear charge
  • The shielding or screening effect of inner electrons
  • Nature of electronic configuration

iii. Variation of ionization energy down the group: On moving down the group, the ionization enthalpy decreases. This is because electron is to be removed from the larger valence shell. Screening due to core electrons goes on increasing and the effective nuclear charge decreases down the group. As a result, the removal of the outer electron becomes easier down the group.

iv. Variation of ionization energy across a period: The screening effect is the same while the effective nuclear charge increases across a period. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period. Therefore, the alkali metal shows the lowest first ionization enthalpy while the inert gas shows the highest first ionization enthalpy across a period.

Note: First ionization enthalpy values of elements of group 1.
Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table 3
Note: First ionization enthalpy values of elements of period 2.
Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table 4

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Question I.
How the atomic size vary in a group and across a period? Explain with suitable example.
Answer:
i. Variation in atomic size down the group:
a. As we move down the group from top to bottom in the periodic table, the atomic size increases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases but simultaneously the number of shells in the atoms also increases.
c. Asa result, the effective nuclear charge decreases due to increase in the size of the atom and shielding effect increases down the group. Thus, the valence electrons experience less attractive force from nucleus and are held less tightly.
d. Hence, the atomic size increases in a group from top to bottom.

e. g.

  • In group 1, as we move from top to bottom i.e., from Li to Cs, a new shell gets added in the atom of the elements and the electrons are added in this new shell.
  • As a result of this, the effective nuclear charge goes on decreasing and screening effect goes on increasing down a group.
  • Therefore, the atomic size is the largest for Cs and is the smallest for Li in group 1.

[Note: Atomic radii of Li and Cs are 152 pm and 262 pm respectively.]

ii. Variation in atomic size across a period:
a. As we move across a period from left to right in the periodic table, the atomic size of an element decreases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.
c. Therefore, as we move across a period, the effective nuclear charge increases but screening effect caused by the core electrons remains the same.
d. As a result of this, attraction between the nucleus and the valence electrons increases. Therefore, valence electrons are more tightly bound and hence, the atomic radius goes on decreasing along a period resulting in decrease in atomic size.

e. g.

  • In the second period, as we move from left towards right i.e., from Li to F, the electrons are added in the second shell of all the elements in second period (except noble gas Ne).
  • As a result of this, the effective nuclear charge goes on increasing from Li to F, however, screening effect remains the same.
  • Therefore, the atomic size is the largest for Li (alkali metal) and is the smallest for F (halogen).

[Note: Atomic radii of Li and F are 152 pm and 64 pm respectively.]

Question J.
Give reasons.
a. Alkali metals have low ionization energies.
b. Inert gases have exceptionally high ionization energies.
c. Fluorine has less electron affinity than chlorine.
d. Noble gases possess relatively large atomic size.
Answer:
a. i. Across a period, the screening effect is the same while the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Since the alkali metals are present in the group 1 of the modem periodic table, they have low ionization energies.

b. i. Across a period, the screening effect is the same and the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Inert gases are present on the extreme right of the periodic table i.e., in group 18. Also, inert gases have stable electronic configurations i.e., complete octet or duplet. Due to this, they are extremely stable and it is very difficult to remove electrons from their valence shell.
Hence, inert gases have exceptionally high ionization potential.

c. The less electron affinity of fluorine is due to its smaller size. Adding an electron to the 2p orbital in fluorine leads to a greater repulsion than adding an electron to the larger 3p orbital of chlorine.
Hence, fluorine has less electron affinity than chlorine.

d. i. Noble gases have completely filled valence shell i.e., complete octet (except He with complete duplet).
ii. Since their valence shell contains eight electrons, they experience greater electronic repulsion and this results in increased atomic size (atomic radii) of the noble gas elements.
Hence, noble gases possess

Question K.
Consider the oxides Li2O, CO2, B2O3.
a. Which oxide would you expect to be the most basic?
b. Which oxide would be the most acidic?
c. Give the formula of an amphoteric oxide.
Answer:
a. Li2O is the most basic oxide.
b. CO2 is the most acidic oxide.
c. Formula of an amphoteric oxide: Al2O3.
[Note: Both B2O3 and CO2 are acidic oxides. But CO2 is more acidic oxide as compared to B2O3. Hence, CO2 is most acidic oxide amongst the given.]

Activity :

Question 1.
Prepare a wall mounting chart of the modern periodic table.
Answer:
Students can scan the adjacent Q.R. Code to visualise the modern periodic table and are expected to prepare the chart on their own.
Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table 5

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

11th Chemistry Digest Chapter 7 Modern Periodic Table Intext Questions and Answers

Can you recall? (Textbook Page No. 93)

Question 1.
What was the basis of classification of elements before the knowledge of electronic structure of atom?
Answer:
Elements were classified on the basis of their physical properties before the knowledge of electronic structure of atom.

Question 2.
Name the scientists who made the classification of elements in the nineteenth century.
Answer:
Dmitri Mendeleev, John Newlands and Johann Doberiener were the scientists who made the classification of elements based on their atomic mass in the nineteenth century.

Question 3.
What is Mendeleev’s periodic law?
Answer:
Mendeleev’s periodic law: “The physical and chemical properties of elements are the periodic function of their atomic masses

Question 4.
How many elements are discovered until now?
Answer:
Including manmade elements, total 118 elements are discovered until now.

Question 5.
How many horizontal rows and vertical columns are present in the modern periodic table?
Answer:
The modem periodic table consists of seven horizontal rows called periods numbered from 1 to 7 and eighteen vertical columns called groups numbered from 1 to 18.

Just think. (Textbook Page No. 93)

Question 1.
How many days pass between two successive full moon nights?
Answer:
29.5 days i.e., approximately 30 days pass between two successive full moon nights.

Question 2.
What type of motion does a pendulum exhibit?
Answer:
A pendulum exhibits periodic motion since it traces the same path after regular interval of time.

Question 3.
Give some other examples of periodic events.
Answer:
Following are some other examples of periodic events:

  • Motion of earth around the sun.
  • Rotation of earth around its own axis.
  • Day and night.

Maharashtra Board Class 11 Chemistry Solutions Chapter 7 Modern Periodic Table

Can you recall? (Textbook Page No. 95)

Question i.
What does the principal quantum number ‘n’ and azimuthal quantum number ‘l’ of an electron belonging to an atom represent?
Answer:
The principal quantum number ‘n’ represents the outermost or valence shell of an element (which corresponds to period number) while azimuthal quantum number ‘l’ constitutes a subshell belonging to the shell for the given ‘n’.

Question ii.
Which principle is followed in the distribution of electrons in an atom?
Answer:
The distribution of electrons in an atom is according to the following three principles:

  1. Aufbau principle
  2. Pauli’s exclusion principle
  3. Hund’s rule of maximum multiplicity

[Note: According to aufbau principle, electrons are filled in the subshells in the increasing order of their energies which follows the following order: s < p < d < f.]

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 5 Gravitation Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 5 Gravitation

1. Choose the correct option.

Question 1.
The value of acceleration due to gravity is maximum at
(A) the equator of the Earth
(B) the centre of the Earth.
(C) the pole of the Earth.
(D) slightly above the surface of the Earth.
Answer:
(C) the pole of the Earth.

Question 2.
The weight of a particle at the centre of the Earth is _________
(A) infinite.
(B) zero.
(C) same as that at other places.
(D) greater than at the poles.
Answer:
(B) zero.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 3.
The gravitational potential due to the Earth is minimum at
(A) the centre of the Earth.
(B) the surface of the Earth.
(C) a points inside the Earth but not at its centre.
(D) infinite distance.
Answer:
(A) the centre of the Earth.

Question 4.
The binding energy of a satellite revolving around planet in a circular orbit is 3 × 109 J. Its kinetic energy is _________
(A) 6 × 109 J
(B) -3 × 109 J
(C) -6 × 10+9 J
(D) 3 × 10+9J
Answer:
(D) 3 × 10+9J

2. Answer the following questions.

Question 1.
State Kepler’s law equal of area.
Answer:
The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.

Question 2.
State Kepler’s law of period.
Answer:
The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.

Question 3.
What are the dimensions of the universal gravitational constant?
Answer:
The dimensions of universal gravitational constant are: [L3M-1T-2].

Question 4.
Define binding energy of a satellite.
Answer:
The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.

Question 5.
What do you mean by geostationary satellite?
Answer:
Some satellites that revolve around the Earth in equatorial plane have same sense of rotation as that of the Earth. The also have the same period of rotation as that of the Earth i.e.. 24 hours. Due to this, these satellites appear stationary from the Earth’s surface and are known as geostationary satellites.

Question 6.
State Newton’s law of gravitation.
Answer:
Statement:
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 7.
Define escape velocity of a satellite.
Answer:
The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (ve) of the body.

Question 8.
What is the variation in acceleration due to gravity with altitude?
Answer:
Variation in acceleration due to gravity due to altitude is given by, gh = g\(\left(\frac{R}{R+h}\right)^{2}\)
where,
gh = acceleration due to gravity of an object placed at h altitude
g = acceleration due to gravity on surface of the Earth
R = radius of the Earth
h = attitude height of the object from the surface of the Earth.
Hence, acceleration due to gravity decreases with increase in altitude.

Question 9.
On which factors does the escape speed of a body from the surface of Earth depend?
Answer:
The escape speed depends only on the mass and radius of the planet.
[Note: Escape velocity does not depend upon the mass of the body]

Question 10.
As we go from one planet to another planet, how will the mass and weight of a body change?
Answer:

  1. As we go from one planet to another, mass of a body remains unaffected.
  2. However, due to change in mass and radius of planet, acceleration due to gravity acting on the body changes as, g ∝ \(\frac{\mathrm{M}}{\mathrm{R}^{2}}\).
    Hence, weight of the body also changes as, W ∝ \(\frac{\mathrm{M}}{\mathrm{R}^{2}}\)

Question 11.
What is periodic time of a geostationary satellite?
Answer:
The periodic time of a geostationary satellite is same as that of the Earth i.e., one day or 24 hours.

Question 12.
State Newton’s law of gravitation and express it in vector form.
Answer:

  1. Statement:
    Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
  2. In vector form, it can be expressed as,
    \(\overrightarrow{\mathrm{F}}_{21}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\left(-\hat{\mathrm{r}}_{21}\right)\)
    where, \(\hat{\mathbf{r}}_{21}\) is the unit vector from m1 to m2.
    The force \(\overrightarrow{\mathrm{F}}_{21}\) is directed from m2 to m1.

Question 13.
What do you mean by gravitational constant? State its SI units.
Answer:

  1. From Newton’s law of gravitation,
    F = G \(\frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
    where, G = constant called universal gravitational constant Its value is 667 X 10-11 N m2/kg2.
  2. G = \(\frac{\mathrm{Fr}^{2}}{\mathrm{~m}_{1} \mathrm{~m}_{2}}\)
    If m1 = m2 = 1 kg, r = 1 m thenF = G.
    Hence, the universal gravitational constant is the force of gravitation between two particles of unit mass separated by unit distance.
  3. Unit: N m2/kg2 in SI system.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 14.
Why is a minimum two stage rocket necessary for launching of a satellite?
Answer:

  1. For the projection of an artificial satellite, it is necessary for the satellite to have a certain velocity.
  2. In a single stage rocket, when the fuel in first stage of rocket is ignited on the surface of the Earth, it raises the satellite vertically.
  3. The velocity of projection of satellite normal to the surface of the Earth is the vertical velocity.
  4. If this vertical velocity is less than the escape velocity (ve), the satellite returns to the Earth’s surface. While, if the vertical velocity is greater than or equal to the escape velocity, the satellite will escape from Earth’s gravitational influence and go to infinity.
  5. Hence, minimum two stage rocket, one to raise the satellite to desired height and another to provide required hori7ontal velocity, is necessary for launching of a satellite.

Question 15.
State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection
Answer:
The path of the satellite depends upon the value of horizontal speed of projection vh relative to critical velocity vc and escape velocity ve.
Case (I) vh < vc:
The orbit of satellite is an ellipse with point of projection as apogee and Earth at one of the foci. During this elliptical path, if the satellite passes through the Earth’s atmosphere. it experiences a nonconservative force of air resistance. As a result it loses energy and spirals down to the Earth.
Case (II) vh = vc:
The satellite moves in a stable circular orbit around the Earth.
Case (III) vc < vh < ve:
The satellite moves in an elliptical orbit round the Earth with the point of projection as perigee.
Case (IV) vh = ve
The satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity.
Case (V) vh > ve:
The satellite escapes from gravitational influence of Earth traversing a hyperbolic path.

3. Answer the following questions in detail.

Question 1.
Derive an expression for critical velocity of a satellite.
Answer:
Expression for critical velocity:

  1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
  2. If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity vc.
  3. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
    ∴ Centripetal force = Gravitational force
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 13
    This is the expression for critical speed at the orbit of radius (R + h).
  4. The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 2.
State any four applications of a communication satellite.
Answer:
Applications of communication satellite:

  1. For the transmission of television and radiowave signals over large areas of Earth’s surface.
  2. For broadcasting telecommunication.
  3. For military purposes.
  4. For navigation surveillance.

Question 3.
Show that acceleration due to gravity at height h above the Earth’s surface is gh = g(\(\frac{R}{R+h}\))2
Answer:
Variation of g due to altitude:

  1. Let,
    R = radius of the Earth,
    M = mass of the Earth.
    g = acceleration due to gravity at the surface of the Earth.
  2. Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is given by,
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 3
  3. The body is taken at height h above the surface of the Earth as shown in figure. The acceleration due to gravity now changes to,
    gh = \(\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}\) …………. (2)
  4. Dividing equation (2) by equation (1), we get,
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 4
    We can rewrite,
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 5
    This expression can be used to calculate the value of g at height h above the surface of the Earth as long as h<< R.

Question 4.
Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 14
vh = horizontal speed of projection
v c = critical velocity
ve = escape velocity

Question 5.
Derive an expression for binding energy of a body at rest on the Earth’s surface.
Answer:

  1. Let,
    M = mass of the Earth
    m = mass of the satellite
    R = radius of the Earth.
  2. Since the satellite is at rest on the Earth, v = 0
    ∴ Kinetic energy of satellite.
    K.E = \(\frac{1}{2}\) mv2 = 0
  3. Gravitational potential at the Earth’s surface
    = – \(\frac{\mathrm{GM}}{\mathrm{R}}\)
    ∴ Potential energy of satellite = Gravitational potential × mass of satellite
    = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
  4. Total energy of sitellite = T.E = P.E + K.E
    ∴ T.E. = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\) + 0 = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
  5. Negative sign in the energy indicates that the satellite is bound to the Earth, due to gravitational force of attraction.
  6. For the satellite to be free form Earth’s gravitational influence, its total energy should become positive. That energy is the binding energy of the satellite at rest on the surface of the Earth.
    ∴ B.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\)

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 6.
Why do astronauts in an orbiting satellite have a feeling of weightlessness?
Answer:

  1. For an astronaut, in a satellite, the net force towards the centre of the Earth will always be, F = mg – N.
    where, N is the normal reaction.
  2. In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.
  3. In this case, the downward acceleration, ad = g, or the satellite (along with the astronaut) is in the state of free fall.
  4. Thus, the net force acting on astronaut will be, F = mg – mad i.e., the apparent weight will be zero, giving the feeling of total weightlessness.

Question 7.
Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth’s surface.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 9

Question 8.
At which place on the Earth’s surface is the gravitational acceleration maximum? Why?
Answer:

  1. Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
  2. At poles, latitude θ = 90°.
    ∴ g’ = g
    i.e., there is no reduction in acceleration due to gravity at poles.
  3. Also, shape of the Earth is actually an ellipsoid, bulged at equator. The polar radius of the Earth is 6356 km which minimum. As g ∝ \(\frac{1}{\mathrm{R}^{2}}\), acceleration due to gravity is maximum at poles i.e., 9.8322 m/s2.

Question 9.
At which place on the Earth surface the gravitational acceleration minimum? Why?
Answer:

  1. Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
  2. At equator, latitude θ = 0°.
    ∴ g’ = g – Rω2
    i.e., the acceleration due to gravity ¡s reduced by amount Rω2(≈ 0.034 m/s2) at equator.
  3. Also, shape of the Earth is actually an ellipsoid, bulged at equator. The equatorial radius of the Earth is 6378 km, which is maximum. As g ∝ \(\frac{1}{\mathrm{R}^{2}}\) acceleration due to gravity is minimum on equator i.e., 9.7804 m/s2.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 10.
Define the binding energy of a satellite. Obtain an expression for binding energy of a satellite revolving around the Earth at certain attitude.
Answer:
The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.
Expression for binding energy of satellite revolving in circular orbit round the Earth:

  1. Consider a satellite of mass m revolving at height h above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy.
  2. Let M be the mass of the Earth, R be the Radius of the Earth, vc be critical velocity of satellite, r = (R + h) be thc radius of the orbit.
  3. Kinetic energy of satellite = \(\frac{1}{2} \mathrm{mv}_{\mathrm{c}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\)
  4. The gravitational potential at a distance r from the centre of the Earth is –\(\frac{\mathrm{GM}}{\mathrm{r}}\)
    ∴ Potential energy of satellite = Gravitational potential × mass of satellite
    = –\(\frac{\mathrm{GMm}}{\mathrm{r}}\)
  5. The total energy of satellite is given as T.E. = KF. + P.E.
    = \(\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}-\frac{\mathrm{GMm}}{\mathrm{r}}=-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\)
  6. Total energy of a circularly orbiting satellite is negative. Negative sign indicates that the satellite is bound to the Earth, due to gravitational force of attraction. For the satellite to be free from the Earth’s gravitational influence its total energy should become zero or positive.
  7. Hence the minimum energy to be supplied to unbind the satellite is +\(\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\). This is the binding energy of a satellite.

Question 11.
Obtain the formula for acceleration due to gravity at the depth ‘d’ below the Earth’s surface.
Answer:

  1. The Earth can be considered to be a sphere made of large number of concentric uniform spherical shells.
  2. When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.
  3. The acceleration due to gravity on the surface of the Earth is, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
  4. Assuming that the density of the Earth is uniform, mass of the Earth is given by
    M = volume x density = \(\frac{4}{3}\) πR3ρ
    ∴ g = \(\frac{\mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}^{2}}\) = \(\frac{4}{3}\) πRρG ………….. (1)
  5. Consider a body at a point P at the depth d below the surface of the Earth as shown in figure.
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 6
    Here the force on a body at P due to outer spherical shell shown by shaded region, cancel out due to symmetry.
    The net force on P is only due to the inner sphere of radius OP = R – d.
  6. Acceleration due to gravity because of this sphere is,
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 7
    This equation gives acceleration due to gravity at depth d below the Earth’s surface.

Question 12.
State Kepler’s three laws of planetary motion.
Answer:

  • All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.
  • The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
  • The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 13.
State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’ Hence show that their ratio is equal to \(\left(\frac{R-d}{R-2 h}\right)\) by assuming that the altitude is very small as compared to the radius of the Earth.
Answer:

  1. For an object at depth d, acceleration due to gravity of the Earth is given by,
    gd = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) ………………. (1)
  2. Also, the acceleration due to gravity at smaller altitude h is given by,
    gh = g(1 – \(\frac{2 \mathrm{~h}}{\mathrm{R}}\)) ……………. (2)
  3. Hence, dividing equation (1) by equation (2),
    we get,
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 8

Question 14.
What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?
Answer:
The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbIt round the Earth is called the critical velocity or orbital velocity (vc).
Expression for critical velocity:

  1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
  2. If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity vc.
  3. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
    ∴ Centripetal force = Gravitational force
    Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 13
    This is the expression for critical speed at the orbit of radius (R + h).
  4. The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.

Question 15.
Define escape speed. Derive an expression for the escape speed of an object from the surface of the earth.
Answer:

  1. The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (ve) of the body.
  2. As the gravitational force due to Earth becomes zero at infinite distance, the object has to reach infinite distance in order to escape.
  3. Let us consider the kinetic and potential energies of an object thrown vertically upwards with escape velocity ve.
  4. On the surface of the Earth,
    K.E.= \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}\)
    P.E. = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
    Total energy = P.E. + K.E.
    ∴ T.E. = \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}\) ………………. (1)
  5. The kinetic energy of the object will go on decreasing with time as it is pulled back by Earth’s gravitational force. It will become zero when it reaches infinity. Thus, at infinite distance from the Earth,
    K.E. = 0
    Also,
    P.E. = –\(\frac{\mathrm{GMm}}{\infty}\) = 0
    ∴ Total energy = P.E. + K.E. = 0
  6. As energy is conserved
    \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}=0\) ……[From(1)]
    or, ve = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 16.
Describe how an artificial satellite using two stage rocket is launched in an orbit around the Earth.
Answer:

  1. Launching of a satellite in an orbit around the Earth cannot take place by use of single stage rocket. It requires minimum two stage rocket.
  2. With the help of first stage of rocket, satellite can be taken to a desired height above the surface of the Earth.
  3. Then the launcher is rotated in horizontal direction i.e.. through 900 using remote control and the first stage of the rocket is detached.
  4. With the help of second stage of rocket, a specific horizontal velocity (vh) is given to satellite so that it can revolve in a circular path around the Earth.
  5. The satellite follows different paths depending upon the horizontal velocity provided to it.

4. Solve the following problems.

Question 1.
At what distance below the surface of the Earth, the acceleration due to gravity decreases by 10% of its value at the surface, given radius of Earth is 6400 km.
Solution:
Given: gd = 90% of g i.e., \(\frac{\mathrm{g}_{\mathrm{d}}}{\mathrm{g}}\) = 0.9,
R = 6400km = 6.4 × 106 m
To find: Distance below the Earth’s surface (d)
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 10
At distance 640 km below the surface of the Earth, value of acceleration due to gravity decreases by 10%.

Question 2.
If the Earth were made of wood, the mass of wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate escape speed from the surface of this Earth.
Solution:
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 12
As, we know that the escape speed from surface of the Earth is 11.2 km/s, Substituting value of ve = 11.2 km/s
Vew = 11.2 × \(\frac{1}{\sqrt{10}}=\frac{11.2}{3.162}\)
= 11.2 × \(\frac{1}{3.162}\)
…………… [Taking square root value]
= antilog {log(1 1.2) –  Log(3.162)}
= antilog {1.0492 – 0.5000}
= antilog {0.5492} = 3.542
∴ Vew = 3.54km/s
The escape velocity from the surface of wooden Earth is 3.54 km/s.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 3.
Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth.
Given:- G = 6.67 × 10-11 Nm2/kg2
R = 6400 km
M = 6 × 1024 kg
Solution:
Given:- m = 2000 kg, h = 3600 km = 3.6 × 106 m,
G = 6.67 × 10-11 Nm2/kg2
R = 6400 km
M = 6 × 1024 kg

To find: i) Kninetic energy (K.E.)
ii) Potential Energy (P.E.)
iii) Total Energy (T.E.)
iv) Binding Energy (B.E.)
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 17
From formula (ii),
P.E. = -2 × 40.02 × 109
= -80.04 × 109 J
From formula (iii),
T.E. = (40.02 × 109) + (-80.02 × 109)
= -40.02 × 109 J
From formula (iv),
B.E.= -(-40.02 × 109)
= 40.02 × 109 J
Kinetic energy of the satellite is 40.02 × 109 J, potential energy is -80.04 × 109 J, total energy is -40.02 × 109 J and binding energy is 40.02 × 109 J.
[Note: Total energy of orbiting satellite is negative.]

Question 4.
Two satellites A and B are revolving around a planet. Their periods of revolution are 1 hour and 8 hours respectively. The radius of orbit of satellite B is 4 × 104 km. find radius of orbit of satellite A .
Solution:
Given: TA = 1 hour, TB = 8 hour,
rB = 4 × 104 km
To find: Radius of orbit of satellite A (rA)
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 16
Radius of orbit of satellite A will be 1 × 104 km.

Question 5.
Find the gravitational force between the Sun and the Earth.
Given Mass of the Sun = 1.99 × 1030 kg
Mass of the Earth = 5.98 × 1024 kg
The average distance between the Earth and the Sun = 1.5 × 1011 m.
Solution:
Given: MS = 1.99 × 1030 kg
ME = 5.98 × 1024 kg, R = 1.5 × 1011 m.
To find: Gravitational force between the Sun and the Earth (F)
Formula: F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
Calculation:As, we know, G = 6.67 × 10-11 N m2/kg2
From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 1
= antilog {(log(6.67) + log( 1.99) + log(5.98) – log(2.25)} × 1021
= antilog {(0.8241) + (0.2989) + (0.7767) – (0.3522)} × 1021
= antilog {1.5475} × 1021
= 35.28 × 1021
= 3.5 × 1022 N
The gravitational force between the Sun and the Earth is 3.5 × 1022 N.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 6.
Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 × 1024 kg, R = 6400 km).
Solution:
Given: h = 300 km = 0.3 × 106 m,
M = 5.98 × 1024 kg,
R = 6400km = 6.4 × 106 m
G = 6.67 × 10-11 Nm2/kg2
To find: Acceleration due to gravity at height (gh)
Formula: gh = \(\frac{G M}{(R+h)^{2}}\)

Calculation: From formula,
gh = \(\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{\left[\left(6.4 \times 10^{6}\right)+\left(0.3 \times 10^{6}\right)\right]^{2}}\)
= \(\frac{6.67 \times 5.98 \times 10^{13}}{(6.7)^{2} \times 10^{12}}\)
6.67 X 10” x 5.98 X iO
= antilog {log(6.67) + log(5.98) – 2log(6.7)} × 10
= antilog{0.8241 + 0.7767 – 2(0.8261)} × 10
= antilog {1.6008 – 1.6522} × 10
= antilog {\(\overline{1}\) .9486} × 10
= 0.8884 × 10 = 8.884 m/s2
Acceleration due to gravity at 300 km will be 8.884 m/s2.

Question 7.
Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface. ME = 5.98 × 1024 kg, R = 6.4 × 106 m.
Solution:
Given: h = 1000 km = 1 × 106 m,
ME = 5.98 × 1024 kg, R = 6.4 × 106 m,
G = 6.67 × 10-11 N m2/kg2
To find: Speed of satellite (vc)
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 15
Speed of the satellite at height 1000 km is 7.34 × 103 m/s.

Question 8.
Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars = 3.4 × 103 km and its mass is 6.4 × 1023 kg.
Solution:
Given:
M = 6.4 × 1023 kg
R = 3.4 × 103 = 3.4 × 106 m,
To find: Acceleration due to gravity on the surface of the Mars (gM)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: As, G = 6.67 × 10-11 N m2/kg2
From formula,
gM = \(\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{\left(3.4 \times 10^{6}\right)^{2}}=\frac{6.67 \times 6.4}{3.4 \times 3.4}\)
= antilog {log(6.67) + log(6.4) – log(3.4) – log(3.4)}
= antilog {(0.8241) + (0.8062) – (0.5315) – (0.53 15)}
= antilog {0.5673}
= 3.693 m/s2
Acceleration due to gravity on the surface of Mars is 3.693 m/s2.

Question 9.
A planet has mass 6.4 × 1024 kg and radius 3.4 × 106 m. Calculate energy required to remove on object of mass 800 kg from the surface of the planet to infinity.
Solution:
Given: M = 6.4 × 1024 kg, R = 3.4 × 106 m, m = 800 kg
To find:   Energy required to remove the object from surface of planet to infinity = B.E.
Formula:    B.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Calculation: We know that,
G = 6.67 × 10-11 N m2/kg2
From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 18
= antilog{log(6.67) + log(51.2) – log(3.4)} × 109
= antilog{0.8241 + 1.7093 – 0.5315} × 109
= antilog {2.0019} × 109
= 1.004 × 102 × 109
= 1.004 × 1011 J
Energy required to remove the object from the surface of the planet is 1.004 × 1011 J.
[Note: Answer calculated above ¡s in accordance with retual methods of calculation.]

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 10.
Calculate the value of the universal gravitational constant from the given data. Mass of the Earth = 6 × 1024 kg, Radius of the Earth = 6400 km and the acceleration due to gravity on the surface = 9.8 m/s2
Solution:
Given: M = 6 × 1024 kg,
R = 6400km = 6.4 × 106 m,
g = 9.8 m/s2
To find: Gravitational constant (G)
Formula. g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,
G = \(\frac{\mathrm{gR}^{2}}{\mathrm{M}}\)
G = \(\frac{9.8 \times\left(6.4 \times 10^{6}\right)^{2}}{6 \times 10^{24}}=\frac{401.4 \times 10^{12}}{6 \times 10^{24}}\)
∴ G = 6.69 × 10-11 N m2/kg2
The value of gravitational constant is 6.69 × 10-11 N m2/kg2.

Question 11.
A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 1/7 times the mass of the Earth and radius twice that of the Earth’s radius.
Solution:
Given: WE = 5.6 kg-wt.,
\(\frac{\mathrm{M}_{\mathrm{p}}}{\mathrm{M}_{\mathrm{E}}}=\frac{1}{7}, \frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{E}}}\) = 2
To find: Weight of the body on the surface of planet (Wp)
Formula: W = mg = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 2
Weight of the body on the surface of a planet will be 0.2 kg-wt.
[Note: The answer given above is calculated in accordance with textual method considering the given data].

Question 12.
What is the gravitational potential due to the Earth at a point which is at a height of 2RE above the surface of the Earth, Mass of the Earth is 6 × 1024 kg, radius of the Earth = 6400 km and G = 6.67 × 10-11 Nm2 kg-2.
Solution:
Given: M = 6 × 1024 kg,
RE = 6400km = 6.4 × 106 m,
G = 6.67 × 10-11 Nm2/kg2,
h = 2RE
To find: Gravitational potential (V)
Formula: V = – \(\frac{\mathrm{GM}}{\mathrm{r}}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation 11
= -2.08 × 107 J kg-1
Negative sign indicates the attractive nature of gravitational potential.
Gravitational potential due to Earth will be 2.08 × 107 J kg-1 towards the centre of the Earth.
[Note: According lo definition of gravitational potential its SI unit is J/kg.]

11th Physics Digest Chapter 5 Gravitation Intext Questions and Answers

Can you recall? (Textbook Page No. 78)

Question 1.
i) What are Kepler’s laws?
ii)What is the shape of the orbits of planets?
Answer:

  1. The Kepler’s laws are:
    • Kepler’s first law: The orbit of a planet is an ellipse with the Sun at one of the foci.
    • Kepler’s second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
    • Kepler’s third law: The square of orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
  2. The orbits of the planet are elliptical in shape.

Maharashtra Board Class 11 Physics Solutions Chapter 5 Gravitation

Question 2.
When released from certain height why do objects tend to fall vertically downwards?
Answer:
When released from certain height, objects tend to fall vertically downwards because of the gravitational force exerted by the Earth.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 4 Laws of Motion Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 4 Laws of Motion

1. Choose the correct answer.

Question 1.
Consider the following pair of forces of equal magnitude and opposite directions:
(P) Gravitational forces are exerted on each other by two-point masses separated by a distance.
(Q) Couple of forces are used to rotate a water tap.
(R) Gravitational force and normal force are experienced by an object kept on a table.
For which of these pair/pairs the two forces do NOT cancel each other’s translational
effect?
(A) Only P
(B) Only P and Q
(C) Only R
(D) Only Q and R
Answer:
(A) Only P

Question 2.
Consider the following forces: (w) Force due to tension along a string, (x) Normal force given by a surface, (y) Force due to air resistance, and (z) Buoyant force or upthrust given by a fluid.
Which of these are electromagnetic forces?
(A) Only w, y, and z
(B) Only w, x, and y
(C) Only x, y and z
(D) All four.
Answer:
(D) All four.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 3.
At a given instant three-point masses m, 2m, and 3m are equidistant from each other. Consider only the gravitational forces between them. Select correct statement/s for this instance only:
(A) Mass m experiences maximum force.
(B) Mass 2m experiences a maximum force.
(C) Mass 3m experiences a maximum force.
(D) All masses experience a force of the same magnitude.
Answer:
(C) Mass 3m experiences a maximum force.

Question 4.
The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along with the table, which is proportional to the resultant normal force given by the table. Forces F1 and F2 act at the same angle θ with the horizontal and both are just initiating the sliding motion of the block along with the table. Force F1 is a pulling force while the force F2 is a pushing force. F2 > F1, because
(A) Component of F2 adds up to weight to increase the normal reaction.
(B) Component of F1 adds up to weight to increase the normal reaction.
(C) Component of F2 adds up to the opposing force.
(D) Component of F1 adds up to the opposing force.
Answer:
(A) Component of F2 adds up to weight to increase the normal reaction.

Question 5.
A mass 2m moving with some speed is directly approaching another mass m moving with double speed. After some time, they collide with coefficient of restitution 0.5. Ratio of their respective speeds after collision is
(A) 2/3
(B) 3/2
(C) 2
(D) ½
Answer:
(B) 3/2

Question 6.
A uniform rod of mass 2m is held horizontal by two sturdy, practically inextensible vertical strings tied at its ends. A boy of mass 3m hangs himself at one third length of the rod. Ratio of the tension in the string close to the boy to that in the other string is
(A) 2
(B) 1.5
(C) 4/3
(D) 5/3
Answer:
(B) 1.5

Question 7.
Select WRONG statement about centre of mass:
(A) Centre of mass of a ‘C’ shaped uniform rod can never be a point on that rod.
(B) If the line of action of a force passes through the centre of mass, the moment of that force is zero.
(C) Centre of mass of our Earth is not at its geometrical centre.
(D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.
Answer:
(D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.

Question 8.
For which of the following objects will the centre of mass NOT be at their geometrical centre?
(I) An egg
(II) a cylindrical box full of rice
(III) a cubical box containing assorted sweets
(A) Only (I)
(B) Only (I) and (II)
(C) Only (III)
(D) All, (I), (II) and (III).
Answer:
(D) All, (I), (II) and (III).

2. Answer the following questions.

Question 1.
In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write respectively against A, B, C and D.
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 1
Answer:

  1. Force due to tension in string: Electromagnetic (EM) force, reaction force, non-conservative force.
  2. Normal force: Electromagnetic (EM) force, non-conservative force. Reaction force
  3. Frictional force: Electromagnetic (EM) force, reaction force, non-conservative force.
  4. Resistive force offered by air or water for objects moving through it: Electromagnetic (EM) force, non-conservative force.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 2.
In real life objects, never travel with uniform velocity, even on a horizontal surface, unless something is done? Why
is it so? What is to be done?
Answer:

  1. According to Newton’s first law, for a body to achieve uniform velocity, the net force acting on it should be zero.
  2. In real life, a body in motion is constantly being acted upon by resistive or opposing force like friction, in the direction opposite to that of the motion.
  3. To overcome these opposing forces, an additional external force is required. Thus, the net force is not maintained at zero, making it hard to achieve uniform velocity.

Question 3.
For the study of any kind of motion, we never use Newton’s first law of motion directly. Why should it be studied?
Answer:

  1. Newton’s first law shows an equivalence between the ‘state of rest’ and ‘state of uniform motion along a straight line.’
  2. Newton’s first law of motion defines force as a physical quantity that brings about a change in ‘state of rest’ or ‘state of .uniform motion along a straight line’ of a body.
  3. Newton’s first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line. Due to all these reasons, Newton’s first law should be studied.

Question 4.
Are there any situations in which we cannot apply Newton’s laws of motion? Is there any alternative for it?
Answer:

  1. Limitation: Newton’s laws of motion cannot be applied for objects moving in non-inertial (accelerated) frame of reference.
    Alternative solution: For non-inertial (accelerated) frame of reference, pseudo force needs to be considered along with all the other forces.
  2. Limitation: Newton’s laws of motion are applicable to point objects and rigid bodies. Alternative solution: Body needs to be approximated as a particle as the laws can be applied to individual particles in a rigid body and then summed up over the body.
  3. Limitation: Newton’s laws of motion cannot be applied for objects moving with speeds comparable to that of light.
    Alternative solution: Einstein’s special theory of relativity has to be used.
  4. Limitation: Newton’s laws of motion cannot be applied for studying the behaviour and interactions of objects having atomic or molecular sizes.
    Alternative solution: Quantum mechanics has to be used.

Question 5.
You are inside a closed capsule from where you are not able to see anything about the outside world. Suddenly you feel that you are pushed towards your right. Can you explain the possible cause (s)? Is it a feeling or a reality? Give at least one more situation like this.
Answer:

  1. In a capsule, if we suddenly feel a push towards the right it is because the capsule is in motion and taking a turn towards the left.
  2. The push towards the right is a feeling. In reality, when the capsule is beginning its turning motion towards the left, we continue in a straight line.
  3. This happens because we try to maintain our direction of motion while the capsule takes a turn towards the left.
  4. An external force is required to change our direction of motion. In accordance with one of the inferences from Newton’s first law of motion, in the absence of any external force, we continue to move in a straight line at constant speed and feel the sudden push in the direction opposite to the motion of the capsule.
  5. Example: While travelling by bus, when the bus takes a sudden turn we feel the push in the opposite direction.

Question 6.
Among the four fundamental forces, only one force governs your daily life almost entirely. Justify the statement by stating that force.
Answer:

  1. Electromagnetic force is the attractive and repulsive force between electrically charged particles.
  2. Since electromagnetic force is much stronger than the gravitational force, it dominates all the phenomena on atomic and molecular scales.
  3. Majority of the forces experienced in our daily life like friction, normal reaction, tension in strings, elastic forces, viscosity etc. are electromagnetic in nature.
  4. The structure of atoms and molecules, the dynamics of chemical reactions etc. are governed by electromagnetic forces.

Thus, out of the four fundamental forces, electromagnetic force governs our daily life almost entirely.

Question 7.
Find the odd man out:
(i) Force responsible for a string to become taut on stretching
(ii) Weight of an object
(iii) The force due to which we can hold an object in hand.
Answer:
Weight of an object.
Reason: Weight of an object (force due to gravity) is a non-contact force while force responsible for a string to become taut (tension force) and force due to which we can hold an object in hand (normal force) are contact forces.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 8.
You are sitting next to your friend on ground. Is there any gravitational force of attraction between you two? If so, why are you not coming together naturally? Is any force other than the gravitational force of the earth coming in picture?
Answer:

  1. Yes, there exists a gravitational force between me and my friend sitting beside each other.
  2. The gravitational force between any two objects is given by, \(\overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) Where,
    G = universal gravitational constant, m1 and m2 = mass of the two objects, r = distance between centres of the two objects
  3. Thus, me and my friend attract each other. But due to our small masses, we exert a force on each other, which is too small as compared to the gravitational force of the earth. Hence, me and my friend don’t move towards each other.
  4. Apart from gravitational force of the earth, there is the normal force and frictional force acting on both me and my friend.

In Chapter 5, you will study about force of gravitation in detail.

Question 9.
Distinguish between:
(A) Real and pseudo forces,
(B) Conservative and non-conservative forces,
(C) Contact and non-contact forces,
(D) Inertial and non-inertial frames of reference.
Answer:
(A) Real and pseudo forces,

No Real force Pseudo Force
i. A force which is produced due to interaction between the objects is called real force. A pseudo force is one which arises due to the acceleration of the observer’s frame of reference.
ii. Real forces obey Newton’s laws of motion. Pseudo forces do not obey Newton’s laws of motion.
iii. Real forces are one of the four fundamental forces. Pseudo forces are not among any of the four fundamental forces.
Example: The earth revolves around the sun in circular path due to gravitational force of attraction between the sun and the earth. Example: Bus is moving with an acceleration (a) on a straight road in forward direction, a person of mass ‘m’ experiences a backward pseudo force of magnitude ‘ma’.

(B) Conservative and non-conservative forces,

No Conservative Non-conservative forces
i. If work done by or against a force is independent’ of the actual path, the force is said to be a conservative force. If work done by or against a force is dependent of the actual path, the force is said to be a non- conservative force.
ii. During work done by a conservative force, the mechanical energy is conserved. During work done by a non­ conservative force, the mechanical energy may not be conserved.
iii. Work done is completely recoverable. Work done is not recoverable.
Example:
gravitational force, magnetic force etc.
Example:
Frictional force, air drag etc.

(C) Contact and non-contact forces,

No Contact forces Non-contact forces
i. The forces experienced by a body due to physical contact are called contact forces. The forces experienced by a body without any physical contact are called non-contact forces.
ii. Example: gravitational force, electrostatic force, magnetostatic force etc. Example: Frictional force, force exerted due to collision, normal reaction etc.

(D) Inertial and non-inertial frames of reference.

No. Inertial frame of reference Non-inertial frame of reference
i The body moves with a constant velocity (can be zero). The body moves with variable velocity.
ii. Newton’s laws are Newton’s laws are
iii. The body does not accelerate. The body undergoes acceleration.
iv. In this frame, force acting on a body is a real force. The acceleration of the frame gives rise to a pseudo force.
Example: A rocket in inter-galactic space (gravity free space between galaxies) with all its engine shut. Example: If a car just starts its motion from rest, then during the time of acceleration the car will be in a non- inertial frame of reference.

Question 10.
State the formula for calculating work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly. Is there any mathematical way out for it? Explain.
Answer:

  1. Suppose a constant force \(\overrightarrow{\mathrm{F}}\) acting on a body produces a displacement \(\overrightarrow{\mathrm{S}}\) in the body along the positive X-direction. Then the work done by the force is given as,
    W = F.s cos θ
    Where θ is the angle between the applied force and displacement.
  2. If displacement is in the direction of the force applied, θ = 0°
    W = \(\overrightarrow{\mathrm{F}}\).\(\overrightarrow{\mathrm{s}}\)

Conditions/limitations for application of work formula:

  1. The formula for work done is applicable only if both force \(\overrightarrow{\mathrm{F}}\) and displacement \(\overrightarrow{\mathrm{s}}\) are constant and finite i.e., it cannot be applied when the force is variable.
  2. The formula is not applicable in several real- life situations like lifting an object through several thousand kilometres since the gravitational force is not constant. It is not applicable to viscous forces like fluid resistance as they depend upon speed and thus are often not constant with time.
  3. The method of integration has to be applied to find the work done by a variable force.

Integral method to find work done by a variable force:

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 2

  1. Let the force vary non-linearly in magnitude between the points A and B as shown in figure (a).
  2. In order to calculate the total work done during the displacement from s1 to s2, we need to use integration. For integration, we need to divide the displacement into large numbers of infinitesimal (infinitely small) displacements.
  3. Let at P1, the magnitude of force be F = P1P1‘. Due to this force, the body displaces through infinitesimally small displacement ds, in the direction of force.
    It moves from P1 to P2.
    ∴ \(\mathrm{d} \overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{P}_{1} \mathrm{P}_{2}}\)
  4. But direction of force and displacement are same, we have
    \(\mathrm{d} \overrightarrow{\mathrm{s}}=\mathrm{P}_{1}{ }^{\prime} \mathrm{P}_{2}^{\prime}\)
  5. \(\mathrm{d} \overrightarrow{\mathrm{s}}\) is so small that the force F is practically constant for the displacement. As the force is constant, the area of the strip \(\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}\) is the work done dW for this displacement.
  6. Hence, small work done between P1 to P2 is dW and is given by
    dW = \(\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}\) = \(\mathrm{P}_{1} \mathrm{P}_{1}^{\prime} \times \mathrm{P}_{1}^{\prime} \mathrm{P}_{2}^{\prime}\)
    = Area of the strip P1P2P2‘P1‘.
  7. The total work done can be found out by dividing the portion AB into small strips like P1P2P2‘P1‘ and taking sum of all the areas of the strips.
    ∴ W = \int_{s_{1}}^{s 2} \vec{F} \cdot d \vec{s}=\text { Area } A B B^{\prime} A^{\prime}\(\)
  8. Method of integration is applicable if the exact way of variation in \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{s}}\) is known and that function is integrable.
  9. The work done by the non-linear variable force is represented by the area under the portion of force-displacement graph.
  10. Similarly, in case of a linear variable force, the area under the curve from s1 to s2 (trapezium APQB) gives total work done W [figure (b)].
    Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 3

Question 11.
Justify the statement, “Work and energy are the two sides of a coin”.
Answer:

  1. Work and energy both are scalar quantities.
  2. Work and energy both have the same dimensions i.e., [M1L2T-2].
  3. Work and energy both have the same units i.e., SI unit: joule and CGS unit: erg.
  4. Energy refers to the total amount of work a body can do.
  5. A body capable of doing more work possesses more energy and vice versa.
  6. Work done on a body by a conservative force is equal to the change in its kinetic energy.

Thus, work and energy are the two sides of the same

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 12.
From the terrace of a building of height H, you dropped a ball of mass m. It reached the ground with speed v. Is the relation mgH = \(\frac{1}{2} m \mathrm{v}^{2}\) applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped?
Answer:

  1. Let the ball dropped from the terrace of a building of height h have mass m. During free fall, the ball is acted upon by gravity (accelerating conservative force).
  2. While coming down, the work that is done is equal to the decrease in the potential energy.
  3. This work done however is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance (retarding non-conservative force). This part of energy appears in some other forms such as heat, sound, etc.
  4. Thus, in this case of an accelerating conservative force along with a retarding non-conservative force, the work-energy theorem is given as, Decrease in the gravitational
    P.E. = Increase in the kinetic energy + work done against non-conservative forces.
  5. Thus, the relation mgh = \(\frac{1}{2} \mathrm{mv}^{2}[latex] is not applicable when non-conservative forces are considered. The part of the energy converted to heat, sound etc also needs to be added to the equation,
  6. The ball will not bounce to the same height from where it was dropped due to the loss in kinetic energy during the collision making it an inelastic collision.

Question 13.
State the law of conservation of linear momentum. It is a consequence of which law? Given an example from our daily life for conservation of momentum. Does it hold good during burst of a cracker?
Answer:

  1. Statement: The total momentum of an isolated system is conserved during any interaction.
  2. The law of conservation of linear momentum is a consequence of Newton’s second law of motion, (in combination with Newton’s third law)
  3. Example: When a nail is driven into a wall by striking it with a hammer, the hammer is seen to rebound after striking the nail. This is because the hammer imparts a certain amount of momentum to the nail and the nail imparts an equal and opposite amount of momentum to the hammer.
    Linear momentum conservation during the burst of a cracker:

    • The law of conservation of linear momentum holds good during bursting of a cracker.
    • When a cracker is at rest before explosion, the linear momentum of the cracker is zero.
    • When cracker explodes into number of pieces, scattered in different directions, the vector sum of linear momentum of these pieces is also zero. This is as per the law of conservation of linear momentum.

Question 14.
Define coefficient of restitution and obtain its value for an elastic collision and a perfectly inelastic collision.
Answer:

i. For two colliding bodies, the negative of ratio of relative velocity of separation to relative velocity of approach is called the coefficient of restitution.

ii. Consider an head-on collision of two bodies of masses m1 and m2 with respective initial velocities u1 and u2. As the collision is head on, the colliding masses are along the same line before and after the collision. Relative velocity of approach is given as,
ua = u2 – u1
Let v1 and v2 be their respective velocities after the collision. The relative velocity of recede (or separation) is then vs = v2 – v1
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 4
iii. For a head on elastic collision, According to the principle of conservation of linear momentum,
Total initial momentum = Total final momentum
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 5
iv. For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e., v1 = v2
∴ v1 – v2 = 0
Substituting this in equation (1), e = 0.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 15.
Discuss the following as special cases of elastic collisions and obtain their exact or approximate final velocities in terms
of their initial velocities.
(i) Colliding bodies are identical.
(ii) A very heavy object collides on a lighter object, initially at rest.
(iii) A very light object collides on a comparatively much massive object, initially at rest.
Answer:
The final velocities after a head-on elastic collision is given as,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 6
i. Colliding bodies are identical
If m1 = m2, then v1 = u2 and v2 = u1
Thus, objects will exchange their velocities after head on elastic collision.

ii. A very heavy object collides with a lighter object, initially at rest.
Let m1 be the mass of the heavier body and m2 be the mass of the lighter body i.e., m1 >> m2; lighter particle is at rest i.e., u2 = 0 then,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 7
i.e., the heavier colliding body is left unaffected and the lighter body which is struck, travels with double the speed of the massive striking body.

iii. A very light object collides on a comparatively much massive object, initially at rest.
If m1 is mass of a light body and m2 is mass of heavy body i.e., m1 << m2 and u2 = 0. Thus, m1 can be neglected.
Hence v1 ≅ -u1, and v2 ≅ 0.
i.e., the tiny (lighter) object rebounds with same speed while the massive object is unaffected.

Question 16.
A bullet of mass m1 travelling with a velocity u strikes a stationary wooden block of mass m2 and gets embedded into it. Determine the expression for loss in the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss?
Answer:

  1. A bullet of mass m1 travelling with a velocity u, striking a stationary wooden block of mass m2 and getting embedded into it is a case of perfectly inelastic collision.
  2. In a perfectly inelastic collision, although there is a loss in kinetic energy, the principle of conservation of energy is not violated as the total energy of the system is conserved.

Loss in the kinetic energy during a perfectly inelastic head on collision:

  1. Let two bodies A and B of masses m1 and m2 move with initial velocity [latex]\overrightarrow{\mathrm{u}_{1}}\), and \(\overrightarrow{\mathrm{u}_{2}}\) respectively such that particle A collides head- on with particle B i.e., u1 > u2.
  2. If the collision is perfectly inelastic, the particles stick together and move with a common velocity \(\overrightarrow{\mathbf{V}}\) after the collision along the same straight line.
    loss in kinetic energy = total initial kinetic energy – total final kinetic energy,
  3. By the law of conservation of momentum, m1u1 + m2u2 = (m1 + m2) v
    ∴ v = \(\frac{\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
  4. Loss of Kinetic energy,
    Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 8
  5. Both the masses and the term (u1 – u2)2 are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as e = 0, the loss is maximum.

Question 17.
One of the effects of a force is to change the momentum. Define the quantity related to this and explain it for a variable
force. Usually when do we define it instead of using the force?
Answer:

  1. Impulse is the quantity related to change in momentum.
  2. Impulse is defined as the change of momentum of an object when the object is acted upon by a force for a given time interval.

Need to define impulse:

  1. In cases when time for which an appreciable force acting on an object is extremely small, it becomes difficult to measure the force and time independently.
  2. In such cases, however, the effect of the force i.e, the change in momentum due to the force is noticeable and can be measured.
  3. For such cases, it is convenient to define impulse itself as a physical quantity.
  4. Example: Hitting a ball with a bat, giving a kick to a foot-ball, hammering a nail, bouncing a ball from a hard surface, etc.

Impulse for a variable force:

  1. Consider the collision between a bat and ball. The variation of the force as a function of time is shown below. The force axis is starting from zero.
  2. From the graph, it can be seen that the force is zero before the impact. It rises to a maximum during the impact and decreases to zero after the impact.
  3. The shaded area or the area under the curve of the force -time graph gives the product of force against the corresponding time (∆t) which is the impulse of the force.
    Area of ABCDE = F. ∆t = impulse of force
  4. For a constant force, the area under the curve is a rectangle.
  5. In case of a softer tennis ball, the collision time becomes larger and the maximum force becomes less keeping the area under curve of the (F – t) graph same.
    Area of ABCDE = Area of PQRST

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 9
In chapter 3, you have studied the concept of using area under the curve.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 18.
While rotating an object or while opening a door or a water tap we apply a force or forces. Under which conditions is this process easy for us? Why? Define the vector quantity concerned. How does it differ for a single force and for two opposite forces with different lines of action?
Answer:

  1. Opening a door can be done with ease if the force applied is:
    • proportional to the mass of the object
    • far away from the axis of rotation and the direction of force is perpendicular to the line joining the axis of rotation with the point of application of force.
  2. This is because, the rotational ability of a force depends not only upon the magnitude and direction of force but also on the point where the force acts with respect to the axis of rotation.
  3. Rotating an object like a water tap can be done with ease if the two forces are equal in magnitude but opposite in direction are applied along different lines of action.
  4. The ability of a force to produce rotational motion is measured by its turning effect called ‘moment of force’ or ‘torque’.
  5. However, a moment of couple or rotational effect of a couple is also called torque.
  6. For differences in the two vector quantities.
No. Moment of a force Moment of a couple
i. Moment of a force is given as, \(\vec{\tau}=\vec{r} \times \vec{F}\) Moment of a couple is given as, \(\vec{\tau}=\vec{r}_{12} \times \vec{F}_{1}=\vec{r}_{21} \times \vec{F}_{2}\)
ii. It depends upon the axis of rotation and the point of application of the force. It depends only upon the two forces, i.e., it is independent of the axis of rotation or the points of application of forces.
iii. It can produce translational acceleration also, if the axis of rotation is not fixed or if friction is not enough. Does not produce any translational acceleration, but produces only rotational or angular acceleration.
iv. Its rotational effect can be balanced by a proper single force or by a proper couple. Its rotational effect can be balanced only by another couple of equal and opposite torque.

Question 19.
Why is the moment of a couple independent of the axis of rotation even if the axis is fixed?
Answer:

  1. Consider a rectangular sheet free to rotate only about a fixed axis of rotation, perpendicular to the plane.
  2. A couple of forces \(\overrightarrow{\mathrm{F}}\) and –\(\overrightarrow{\mathrm{F}}\) is acting on the sheet at two different locations.
  3. Consider the torque of the couple as two torques due to individual forces causing rotation about the axis of rotation.
  4. Case 1: The axis of rotation is between the lines of action of the two forces constituting the couple. Let x and y be the perpendicular distances of the axis of rotation from the forces \(\overrightarrow{\mathrm{F}}\) and –\(\overrightarrow{\mathrm{F}}\) respectively.
    In this case, the pair of forces cause anticlockwise rotation. As a result, the direction of individual torques due to the two forces is the same.
    Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 10
  5. Case 2: Lines of action of both the forces are on the same side of the axis of rotation. Let q and p be the perpendicular distances of the axis of rotation from the forces \(\overrightarrow{\mathrm{F}}\) and –\(\overrightarrow{\mathrm{F}}\)

Question 20.
Explain balancing or mechanical equilibrium. Linear velocity of a rotating fan as a whole is generally zero. Is it in
mechanical equilibrium? Justify your answer.
Answer:

  1. The state in which the momentum of a system is constant in the absence of an external unbalanced force is called mechanical equilibrium.
  2. A particle is said to be in mechanical equilibrium, if no net force is acting upon it.
  3. In case of a system of bodies to be in mechanical equilibrium, the net force acting on any part of the system should be zero i.e., the velocity or linear momentum of all parts of the system must be constant or zero. There should be no acceleration in any part of the system.
  4. Mathematically, for a system in mechanical equilibrium, \(\sum \vec{F}\) = 0.
  5. In case of rotating fan, if linear velocity is zero, then the linear momentum is zero. That means there is no net force acting on the fan. Hence, the fan is in mechanical equilibrium.

Question 21.
Why do we need to know the centre of mass of an object? For which objects, its position may differ from that of the centre of gravity?
Use g = 10 m s-2, unless, otherwise stated.
Answer:

  1. Centre of mass of an object allows us to apply Newton’s laws of motion to finite objects (objects of measurable size) by considering these objects as point objects.
  2. For objects in non-uniform gravitational field or whose size is comparable to that of the Earth (size at least few thousand km), the position of centre of mass will differ than that of centre of gravity.

3. Solve the following problems.

Question 1.
A truck of mass 5 ton is travelling on a horizontal road with 36 km hr-1 stops on travelling 1 km after its engine fails suddenly. What fraction of its weight is the frictional force exerted by the road? If we assume that the story repeats for a car of mass 1 ton i.e., can moving with same speed stops in similar distance same how much will the fraction be?
[Ans: \(\frac{1}{200}\) in the both]
Solution:
Given: mtruck = 5 ton = 5000 kg,
mcar = 1 ton = 1000 kg,
u = 36km/hr = 10 m/s,
v = 0 m/s, s = 1 km = 1000 m
To find: Ratio of force of friction to the weight of vehicle
Formulae:
i. v2 = u2 + 2as
ii. F = ma
Calculation: From formula (i),
2 × atruck × s = v2 – u2
∴ 2 × atruck × 1000 = 02 – 102
∴ 200atruck = -100
∴ atruck = -0.05 m/s2
Negative sign indicates that velocity is decreasing
From formula (ii),
Ftruck = mtruck × atruck = 5000 × 0.05
= 250 N
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 11
Answer:
The frictional force acting on both the truck and the car is \(\frac{1}{200}\) of their weight.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 2.
A lighter object A and a heavier object B are initially at rest. Both are imparted the same linear momentum. Which will start with greater kinetic energy: A or B or both will start with the same energy?
[Ans: A]
Solution:

  1. Let m1 and m2 be the masses of light object A and heavy object B and v1 and v2 be their respective velocities.
  2. Since both are imparted with the same linear momentum,
    m1 v1 = m2 v2
  3. Kinetic energy of the lighter object A
    Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 12
  4. As m1 < m2, therefore K.E.A > K.E.B, i.e, the lighter body A has more kinetic energy.

Question 3.
As i was standing on a weighing machine inside a lift it recorded 50 kg wt. Suddenly for few seconds it recorded 45 kg wt. What must have happened during that time? Explain with complete numerical analysis. [Ans: Lift must be coming down with acceleration \(\frac{\mathrm{g}}{10}\) = 1 ms-2]
Solution:
The weight recorded by weighing machine is always apparent weight and a measure of reaction force acting on the person. As the apparent weight (45 kg-wt) in this case is less than actual weight (50 kg-wt) the lift must be accelerated downwards during that time.

Numerical Analysis

  1. Weight on the weighing machine inside the lift is recorded as 50 kg-wt
    ∴ mg = 50 kg-wt
    Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 13
  2. This weight acts on the weighing machine which offers a reaction R given by the reading of the weighing machine
    ∴ R = 45kg-wt = \(\frac{9}{10}\)mg
  3. The forces acting on person inside lift are as follows:
    • Weight mg downward (exerted by the earth)
    • Normal reaction (R) upward (exerted by the floor)
  4. As, R < mg, the net force is in downward direction and given as,
    mg – R=ma
    But R = \(\frac{9}{10}\)mg.
    ∴ mg – \(\frac{9}{10}\)mg = ma
    ∴ \(\frac{mg}{10}\) = ma
    ∴ a = g/10
    ∴ a = 1 m/s2 (∵ g = 1 m/s2)
  5. Therefore, the elevator must be accelerated downwards with an acceleration of 1 m/s2 at that time.

Question 4.
Figure below shows a block of mass 35 kg resting on a table. The table is so rough that it offers a self adjusting resistive force 10% of the weight of the block for its sliding motion along the table. A 20 kg wt load is attached to the block and is passed over a pulley to hang freely on the left side. On the right side there is a 2 kg wt pan attached to the block and hung freely. Weights of 1 kg wt each, can be added to the pan. Minimum how many and maximum how many such weights can be added into the pan so that the block does not slide along the table? [Ans: Min 15, maximum 21].
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 14
Solution:
Frictional (resistive) force f = 10% (weight)
= \(\frac{10}{100}\) × 35 × 10 = 35N 100
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 15

i. Consider FBD for 20 kg-wt load. Initially, the block kept on the table is moving towards left, because of the movement of block of mass 20 kg in downward direction.
Thus, for block of mass 20 kg,
ma = mg – T1 …. (1)
Consider the forces acting on the block of mass 35 kg in horizontal direction only as shown in figure (b). Thus, the force equation for this block is, m1a = T1 – T2 – f ….(2)
To prevent the block from sliding across the table,
m1a = ma = 0
∴ T1 = mg = 200 N ….[From (1)]
T1 = T2 + f ….[From (2)]
∴ T2 + f = 200
∴ T2 = 200 – 35 = 165 N
Thus, the total force acting on the block from right hand side should be 165 N.
∴ Total mass = 16.5 kg
∴ Minimum weight to be added = 16.5 – 2 = 14.5 kg
≈ 15 weights of 1 kg each

ii. Now, considering motion of the block towards right, the force equations for the masses in the pan and the block of mass 35 kg can be determined from FBD shown
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 16
From figure (c)
m1a = T2 – T1 – f ….(iii)
From figure (d),
m2a = m2g – T2 … .(iv)
To prevent the block of mass 35 kg from sliding across the table, m1a = m2a = 0
From equations (iii) and (iv),
T2 = T1 + f
T2 = m2g
∴ m2g = 200 + 35 = 235 N
∴ The maximum mass required to stop the sliding = 23.5 – 2 = 21.5kg ≈ 21 weights of 1 kg
Answer:
The minimum 15 weights and maximum 21 weights of 1 kg each are required to stop the block from sliding.

Question 5.
Power is rate of doing work or the rate at which energy is supplied to the system. A constant force F is applied to a body
of mass m. Power delivered by the force at time t from the start is proportional to
(a) t
(b) t2
(c) \(\sqrt{t}\)
(d) t0
Derive the expression for power in terms of F, m and t.
[Ans: p = \(\frac{F^{2} t}{m}\), ∴ p ∝ t]
Solution:
Derivation for expression of power:

i. A constant force F is applied to a body of mass (m) initially at rest (u = 0).

ii. We have,
v = u + at
∴ v = 0 + at
∴ v = at …. (1)

iii. Now, power is the rate of doing work,
∴ P = \(\frac{\mathrm{d} \mathrm{W}}{\mathrm{d} \mathrm{t}}\)
∴ P = F. \(\frac{\mathrm{d} \mathrm{s}}{\mathrm{dt}}\) [∵ dW = F. ds]

iv. But \(\frac{\mathrm{d} \mathrm{s}}{\mathrm{dt}}\) = v, the instantaneous velocity of the particle.
∴ P = F.V … (2)

v. According to Newton’s second law,
F = ma … (3)

vi. Substituting equations (1) and (3) in equation (2)
P = (ma) (at)
∴ P = ma2t
∴ P = \(\frac{m^{2} a^{2}}{m}\) × t
∴ P = \(\frac{\mathrm{F}^{2}}{\mathrm{~m}} \mathrm{t}\)

vii. As F and m are constant, therefore, P ∝ t.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 6.
40000 litre of oil of density 0.9 g cc is pumped from an oil tanker ship into a storage tank at 10 m higher level than the ship in half an hour. What should be the power of the pump? [Ans: 2 kW]
Solution:
h = 10 m, ρ = 0.9 g/cc = 900 kg/m3, g = 10 m/s2,
V = 40000 litre = 40000 × 103 × 10-6 m3
= 40 m3
T = 30 min = 1800 s
To find: Power(P)
Formula: P = \(\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{h} \rho \mathrm{gV}}{\mathrm{t}}\)
Calculation: From formula,
P = \(\frac{10 \times 900 \times 10 \times 40}{1800}\)
∴ P = 2000 W
∴ P = 2 kW
Answer:
The power of the pump is 2 kW.

Question 7.
Ten identical masses (m each) are connected one below the other with 10 strings. Holding the topmost string, the system is accelerated upwards with acceleration g/2. What is the tension in the 6th string from the top (Topmost string being the first string)? [Ans: 6 mg]
Solution
Consider the 6th string from the top. The number of masses below the 6th string is 5. Thus, FBD for the 6th mass is given in figure (b).
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 18
Answer:
Tension in the 6th string is 7.5 mg.
[Note: The answer given above is modified considering the correct textual concepts.]

Question 8.
Two galaxies of masses 9 billion solar mass and 4 billion solar mass are 5 million light-years apart. If, the Sun has to cross the line joining them, without being attracted by either of them, through what point it should pass? [Ans: 3 million light-years from the 9 billion solar mass]
Solution:
The Sun can cross the line joining the two galaxies without being attracted by either of them if it passes from a neutral point. A neutral point is a point on the line joining two objects where the effect of gravitational forces acting due to both the objects is nullified.
Given that;
m1 = 9 × 109 Ms
m2 = 4 × 109 Ms
r = 5 × 106 light years
Let the neutral point be at distance x from mi. If sun is present at that point,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 19
Answer: The Sun has to cross the line from a point at a distance 3 million light years from the galaxy of mass 9 billion solar mass.

Question 9.
While decreasing linearly from 5 N to 3 N, a force displaces an object from 3 m to 5 m. Calculate the work done by this force during this displacement. [Ans: 8 N]
Solution:
For a variable force, work done is given by area under the curve of force v/s displacement graph. From given data, graph can be plotted as follows:
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 20
= 8 J
Ans: Work done is 8 J.
[Note: According to the definition of work done, S.J. unit of wõrk done is joule (J)]

Alternate solution:
Work done, w = Area of trapezium ADCB
∴ W = \(\frac{1}{2}\)(AD + CB) × DC
∴ W = 1 (5N + 3N) × (5m – 3m)
= \(\frac{1}{2}\) × 8 × 2 = 8J

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 10.
Variation of a force in a certain region is given by F = 6x2 – 4x – 8. It displaces an object from x = 1 m to x = 2 m in this region. Calculate the amount of work
done. [Ans: Zero]
Solution:
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 21
Answer:
The work done is zero.

Question 11.
A ball of mass 100 g dropped on the ground from 5 m bounces repeatedly. During every bounce 64% of the potential energy is converted into kinetic energy. Calculate the following:
(a) Coefficient of restitution.
(b) Speed with which the ball comes up from the ground after third bounce.
(c) Impulse given by the ball to the ground during this bounce.
(d) Average force exerted by the ground if this impact lasts for 250 ms.
(e) Average pressure exerted by the ball on the ground during this impact if the contact area of the ball is 0.5 cm2.
[Ans: 0.8, 5.12 m/s, 1.152N s, 4.608 N, 9.216 × 104 N/m2]
Solution:
Given that, for every bounce, 64% of the initial energy is converted to final energy.
i. Coefficient of restitution in case of inelastic collision is given by,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 105

ii. From equation (1),
∴ v = – eu
∴ After first bounce,
v1 = – eu
after second bounce,
v2 = -ev1 = -e(-eu)= e2u
and after third bounce,
v3 = – ev2 = – e(e2u) = – e3u
But u = \(\sqrt{2 \mathrm{gh}}\)
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 106

iii. Impulse given by the ball during third bounce, is,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 110

iv. Average force exerted in 250 ms,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 120

v. Average pressure for area
0.5 cm2 = 0.5 × 10-4m2
P = \(\frac{\mathrm{F}}{\mathrm{A}}=\frac{4.608}{0.5 \times 10^{-4}}\) = 9.216 × 104 N/m2

Question 12.
A spring ball of mass 0.5 kg is dropped from some height. On falling freely for 10 s, it explodes into two fragments of
mass ratio 1:2. The lighter fragment continues to travel downwards with speed of 60 m/s. Calculate the kinetic energy supplied during explosion. [Ans: 200 J]
Solution:
m1 + m2 = 0.5 kg, m1 : m2 = 1 : 2,
m1 = \(\frac{1}{6}\) kg,
∴ m2 = \(\frac{1}{3}\) kg
Initially, when the ball is falling freely for 10s,
v = u + at = 0 + 10(10)
∴ v = 100 m/s = u1 = u2
(m1 + m2)v = m1v1 + m2v2
∴ 0.5 × 100 = \(\frac{1}{6}\)(60) + \(\frac{1}{3}\)v2
∴ 50 = 10 + \(\frac{1}{3}\)v2
∴ 40 = \(\frac{1}{3}\)v2
∴ v2 = 120m/s
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 22
∴ K.E. = 200J.
Ans: Kinetic energy supplied is 200 J.

Question 13.
A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m with double speed. After collision, the heavier one travels with the average initial speed of the two. Calculate the coefficient of restitution. [Ans: 0.5]
Solution:
Given: m1 = 2m, m2 = m, u1 = 6 cm/s,
u2 = 2u1 = 12 cm/s,
v1 = \(\frac{\mathrm{u}_{1}+\mathrm{u}_{2}}{2}\) = 9cm/s
To find: Coefficient of restitution(e)
Formulae:

i. m1u1 +m2u2 = m1v1 + m2v2
ii. e = \(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
Calculation: From formula (i),
[(2m) × 6] + (m × 12) = (2m × 9) + mv2
∴ v2 = 6cm/s

From formula (ii),
e = \(\frac{6-9}{6-12}\) = \(\frac{-3}{-6}\) = 0.5

Answer: The coefficient of restitution is 0.5

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 14.
A 2 m long wooden plank of mass 20 kg is pivoted (supported from below) at 0.5 m from either end. A person of mass 40 kg starts walking from one of these pivots to the farther end. How far can the person walk before the plank topples? [Ans: 1.25 m]
Solution:
Let the person starts walking from pivot P2 as shown in the figure.
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 100
Assume the person can walk up to distance x from P1 before the plank topples. The plank will topple when the moment exerted by the person about P1 is not balanced by a moment of force due to plank about P2.
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 101
∴ For equilibrium,
40 × x = 20 × 0.5
∴ x = \(\frac{1}{4}\) = 0.25 m
Hence, the total distance walked by the person is 1.25 m.

Question 15.
A 2 m long ladder of mass 10 kg is kept against a wall such that its base is 1.2 m away from the wall. The wall is smooth but the ground is rough. The roughness of the ground is such that it offers a maximum horizontal resistive force (for sliding motion) half that of normal reaction at the point of contact. A monkey of mass 20 kg starts climbing the ladder. How far can it climb along the ladder? How much is the horizontal reaction at the wall? [Ans: 1.5 m, 15 N]
Solution:
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 23
From the figure,
Given that, AC = length of ladder = 2 m
BC= 1.2m
From Pythagoras theorem,
AB = \(\sqrt{\mathrm{AC}^{2}-\mathrm{BC}^{2}}\) = 1.6 m … (i)
Also, ∆ABC ~ ∆DD’C
∴ \(\frac{\mathrm{AB}}{\mathrm{DD}^{\prime}}\) = \(\frac{\mathrm{BC}}{\mathrm{D}^{\prime} \mathrm{C}}\) = \(\frac{\mathrm{AC}}{\mathrm{DC}}\)
∴ \(\frac{1.2}{\mathrm{D}^{\prime} \mathrm{C}}=\frac{2}{1}\)
∴ D’C = 0.6 m … (ii)

The ladder exerts horizontal force \(\overrightarrow{\mathrm{H}}\) on the wall at A and \(\overrightarrow{\mathrm{F}}\) is the force exerted on the ground at C.
As |\(\overrightarrow{\mathrm{F}}\)| = \(|\overrightarrow{\mathrm{H}}|=|\overrightarrow{\mathrm{F}}|=\frac{\mathrm{N}}{2}\) … (iii)

Let monkey climb upto distance x along BC (Horizontal) i.e., CM’ = x .. . .(iv)
Then, the net normal reaction at point C will be, N = 100 + 200 = 300N
From equation (iii),
H = \(\frac{\mathrm{N}}{2}=\frac{300}{2}\) = 150N
By condition of equilibrium, taking moments about C,
(-H × AB) + (W1 × CD’) + (W2 × CM’) + (F × 0)’0
∴ (-150 × 1.6) + (100 × 0.6) + (200 × x) = 0
∴ 60 + 200x = 240
∴ 200x = 180
∴ x = 0.9

From figure, it can be shown that,
∆ABC ~ ∆MM’C
∴ \(\frac{\mathrm{BC}}{\mathrm{CM}^{\prime}}\) = \(\frac{\mathrm{AC}}{\mathrm{CM}^{\prime}}\) ∴ \(\frac{\mathrm{1.2}}{\mathrm{0.9}^{\prime}}\) = \(\frac{\mathrm{2}}{\mathrm{CM}^{\prime}}\)
∴ CM = 1.5 m

Answer:

  1. The monkey can climb upto 1.5 m along the ladder.
  2. The horizontal reaction at wall is 150 N.

Question 16.
Four uniform solid cubes of edges 10 cm, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order. Locate centre of mass of their system. [Ans: 65 cm, 17.7 cm]
Solution:
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 24
The given cubes are arranged as shown in figure. Let one of the comers of smallest cube lie at the origin.
As the cubes are uniform, let their centre of masses lie at their respective centres.
rA \(\equiv\) (5, 5), rB \(\equiv\) (20, 10), rC \(\equiv\) = (45, 15) and rD \(\equiv\) – (80, 20)
Also, masses of the cubes are,
mA = \(l_{\mathrm{A}}^{3} \times \rho=10^{3} \rho\)
mB = (20)3ρ
mC = (30)3ρ
mD = (40)3ρ
As the cubes are uniform, p is same for all of them.
∴ For X – co-ordinate of centre of mass of the system,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 25
Similarly,
Y – co-ordinate of centre of mass of system is,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 26
Answer: Centre of mass of the system is located at point (65 cm, 17.7 cm)

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Question 17.
A uniform solid sphere of radius R has a hole of radius R/2 drilled inside it. One end of the hole is at the centre of the
sphere while the other is at the boundary. Locate centre of mass of the remaining sphere. [Ans: -R/14 ]
Solution:
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 27
Let the centre of the sphere be origin O. Then, r1 be the position vector of centre of mass of uniform solid sphere and r2 be the position vector of centre of mass of the cut-out part of the sphere.
Now, mass of the sphere is given as,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 28
∴ Position vector of centre of mass of remaining part,
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 29
rCM = \(\frac{-\mathbf{R}}{14}\)
(Negative sign indicates the distance is on left side of the origin.)
Ans: Position of centre of mass of remaining sphere \(\frac{-\mathbf{R}}{14}\)

Question 18.
In the following table, every item on the left side can match with any number of items on the right hand side. Select all those.
Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion 30
Answer:
i. Elastic collision: (b)
ii. Inelastic collision: (a), (c), (f), (e)
iii. Perfectly inelastic collision: (d)
iv. Head-on collision: (c), (f)

11th Physics Digest Chapter 4 Laws of Motion Intext Questions and Answers

Can you recall? (Textbook Page No. 47)

Question 1.
What are different types of motions?
Answer:
The various types of motion are linear, uniform linear, non-uniform linear, oscillatory, circular, periodic, and random motion.

Question 2.
What do you mean by kinematical equations and what are they?
Answer:
A set of three equations that analyses rectilinear motion of the uniformly accelerated body and help to predict the position of body are called kinematical equations.

  1. Equation for velocity-time relation: v = u + at
  2. Equation for position-time relation: s = ut + \(\frac{1}{2}\)at2
  3. Position-velocity relation: v2 = u2 + 2as

Can you tell? (Textbook Page No. 48)

Question 1.
Was Aristotle correct? If correct, explain his statement with an illustration.
Answer:
Aristotle was not correct in stating that an external force is required to keep a body in uniform motion.

Question 2.
If wrong, give the correct modified version of his statement.
Answer:
For an uninterrupted motion of a body, an additional external force is required for overcoming opposing/resistive forces.

Maharashtra Board Class 11 Physics Solutions Chapter 4 Laws of Motion

Can you tell? (Textbook Page No. 48)

Question 1.
What is then special about Newton’s first law if it is derivable from Newton’s second law?
Answer:

  1. Newton’s first law shows an equivalence between the ‘state of rest’ and ‘state of uniform motion along a straight line.’
  2. Newton’s first law of motion defines force as a physical quantity that brings about a change in ‘state of rest’ or ‘state of .uniform motion along a straight line’ of a body.
  3. Newton’s first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line.
  4. Due to all these reasons, Newton’s first law should be studied.

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 3 Motion in a Plane Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 3 Motion in a Plane

1. Choose the correct option.

Question 1.
An object thrown from a moving bus is on example of __________
(A) Uniform circular motion
(B) Rectilinear motion
(C) Projectile motion
(D) Motion in one dimension
Answer:
(C) Projectile motion

Question 2.
For a particle having a uniform circular motion, which of the following is constant ____________.
(A) Speed
(B) Acceleration
(C) Velocity
(D) Displacement
Answer:
(A) Speed

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 3.
The bob of a conical pendulum undergoes ___________
(A) Rectilinear motion in horizontal plane
(B) Uniform motion in a horizontal circle
(C) Uniform motion in a vertical circle
(D) Rectilinear motion in vertical circle
Answer:
(B) Uniform motion in a horizontal circle

Question 4.
For uniform acceleration in rectilinear motion which of the following is not correct?
(A) Velocity-time graph is linear
(B) Acceleration is the slope of velocity time graph
(C) The area under the velocity-time graph equals displacement
(D) Velocity-time graph is nonlinear
Answer:
(D) Velocity-time graph is nonlinear

Question 5.
If three particles A, B and C are having velocities \(\overrightarrow{\mathrm{v}}_{A}\), \(\overrightarrow{\mathrm{v}}_{B}\) and \(\overrightarrow{\mathrm{v}}_{C}\) which of the following formula gives the relative velocity of A with respect to B
(A) \(\overrightarrow{\mathrm{v}}_{A}\) + \(\overrightarrow{\mathrm{v}}_{B}\)
(B) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{C}\) + \(\overrightarrow{\mathrm{v}}_{B}\)
(C) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{B}\)
(D) \(\overrightarrow{\mathrm{v}}_{C}\) – \(\overrightarrow{\mathrm{v}}_{A}\)
Answer:
(C) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{B}\)

2. Answer the following questions.

Question 1.
Separate the following in groups of scalar and vectors: velocity, speed, displacement, work done, force, power, energy, acceleration, electric charge, angular velocity.
Answer:
Scalars
Speed, work done, power, energy, electric charge.

Vectors
Velocity, displacement, force, acceleration, angular velocity (pseudo vector).

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 2.
Define average velocity and instantaneous velocity. When are they same?
Answer:
Average velocity:

  1. Average velocity (\(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\)) of an object is the displacement (\(\Delta \overrightarrow{\mathrm{x}}\)) of the object during the time interval (∆t) over which average velocity is being calculated, divided by that time interval.
  2. Average velocity = (\(\frac{\text { Displacement }}{\text { Time interval }}\))
    \(\overrightarrow{\mathrm{V}_{\mathrm{av}}}=\frac{\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}\)
  3. Average velocity is a vector quantity.
  4. Its SI unit is m/s and dimensions are [M0L1T-1]
  5. For example, if the positions of an object are x +4 m and x = +6 m at times t = O and t = 1 minute respectively, the magnitude of its average velocity during that time is Vav = (6 – 4)1(1 – 0) = 2 m per minute and its direction will be along the positive X-axis.
    ∴ \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\) = 2 i m/min
    Where, i = unit vector along X-axis.

Instantaneous velocity:

  1. The instantaneous velocity (\(\overrightarrow{\mathrm{V}}\)) is the limiting value of ¡he average velocity of the object over a small time interval (∆t) around t when the value of lime interval goes to zero.
  2. It is the velocity of an object at a given instant of time.
  3. \(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
    where \(\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) derivative of \(\overrightarrow{\mathrm{x}}\) with respect to t.

In case of uniform rectilinear motion, i.e., when an object is moving with constant velocity along a straight line, the average and instantaneous velocity remain same.

Question 3.
Define free fall.
Answer:
The motion of any object under the influence of gravity alone is called as free fall.

Question 4.
If the motion of an object is described by x = f(t) write formulae for instantaneous velocity and acceleration.
Answer:

  1. Instantaneous velocity of an object is given as,
    \(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
  2. Motion of the object is given as, x = f(t)
  3. The derivative f ‘(f) represents the rate of change of the position f (t) at time t, which is the instantaneous velocity of the object.
    ∴ \(\vec{v}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) = f'(t)
  4. Acceleration is defined as the rate of change of velocity with respect to time.
  5. The second derivative of the position function f “(t) represents the rate of change of velocity i.e., acceleration.
    ∴ \(\overrightarrow{\mathrm{a}}=\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta \mathrm{t}}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = f”(t)

Question 5.
Derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutually perpendicular directions.
Answer:

  1. Consider an object moving in an x-y plane. Let the initial velocity of the object be \(\overrightarrow{\mathrm{u}}\) at t = 0 and its velocity at time t be \(\overrightarrow{\mathrm{v}}\).
  2. As the acceleration is constant, the average acceleration and the instantaneous acceleration will be equal.
    Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 6
    This is the first equation of motion in vector form.
  3. Let the displacement of the object from time t
    = 0 to t be \(\overrightarrow{\mathrm{s}}\)
    For constant acceleration, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{u}}}{2}\)
    Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 7
    This is the second equation of motion in vector form.
  4. Equations (1) and (2) can be resolved into their x and y components so as to get corresponding scalar equations as follows.
    vx = ux + axt ………….. (3)
    vy = uy + ay t …………… (4)
    sx = uxt + \(\frac{1}{2}\) axt2 ………….. (5)
    sy = uyt + \(\frac{1}{2}\) ayt2 ………..(6)
  5. It can be seen that equations (3) and (5) involve only the x components of displacement, velocity and acceleration while equations (4) and (6) involve only the y components of these quantities.
  6. Thus, the motion along the x direction of the object is completely controlled by the x components of velocity and acceleration while that along the y direction is completely controlled by the y components of these quantities.
  7. This shows that the two sets of equations are independent of each other and can be solved independently.

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 6.
Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 2

  1. Consider an object starting from position x = 0 at time t = 0. Let the velocity at time (t = 0) and t be u and v respectively.
  2. The slope of line PQ gives the acceleration. Thus
    ∴ a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}-0}=\frac{\mathrm{v}-\mathbf{u}}{\mathrm{t}}\)
    ∴ v = u + at …………… (1)
    This is the first equation of motion.
  3. The area under the curve in velocity-time graph gives the displacement of the object.
    ∴ s = area of the quadrilateral OPQS = area of rectangle OPRS + area of triangle PQR.
    = ut + \(\frac{1}{2}\) (v – u) t
    But, from equation (1)
    at = v – u
    ∴ s = ut + \(\frac{1}{2}\) at2
    This is the second equation of motion,
  4. The velocity is increasing linearly with time as acceleration is constant. The displacement is given as,
    Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 3
    ∴ s = (v2 – u2) / (2a)
    ∴ v2 – u2 = 2as
    This is the third equation of motion.

Question 7.
Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity \(\vec{u}\) at an angel θ to the horizontal.
Answer:
Expression for range:

  1. Consider a body projected with velocity \(\vec{u}\), at an angle θ of projection from point O in the co-ordinate system of the XY- plane, as shown in figure.
  2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
    Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 9
    ux = u cos θ (Horizontal component)
    uy = u sin θ (Vertical component)
  3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to
    vy = uy + ayt
    with ay = -g and uy = u sinθ
  4. Thus, the components of velocity of the projectile at time t are given by,
    vx = ux = u cos θ
    vy = ux – gt = usin θ – gt
  5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
    s = (u cosθ) t
    sy = (usinθ)t – \(\frac{1}{2}\) gt2
  6. At the highest point, the time of ascent of the projectile is given as,
    tA = \(\frac{u \sin \theta}{g}\) …………..(2)
  7. The total time in air i.e., time of flight is given as, T = 2tA = \(\frac{2u \sin \theta}{g}\) …… (3)
  8. The total horizontal distance travelled by the particle in this time T is given as,
    R = ux ∙ T
    R = u cos θ ∙ (2tA)
    R = u cos θ ∙ \(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) ……………[From (3)]
    R = \(\frac{u^{2}(2 \sin \theta \cdot \cos \theta)}{g}\)
    R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) ………..[∵ sin 2θ = 2sin∙cosθ]
    This is required expression for horizontal range of the projectile.

Expression for maximum height of a projectile:
The maximum height H reached by the projectile is the distance travelled along the vertical (y) direction in time tA.

Substituting sy = H and t = ta in equation (1),
we have,
H = (u sin θ)tA – \(\frac{1}{2}\) gtA2
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 10
This equation represents maximum height of projectile.

Question 8.
Show that the path of a projectile is a parabola.
Answer:

  1. Consider a body projected with velocity initial velocity \(\vec{u}\) , at an angle θ of projection from point O in the co-ordinate system of the XY-plane. as shown in figure.
    Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 8
  2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
    ux = u cos θ (Horizontal component)
    uy = u sin θ (Vertical component)
  3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to,
    vy = uy + ay t
    with ay, = -g and uy = u sinθ
  4. Thus, the components of velocity of the projectile at time t are given by,
    vx = ux = u cosθ
    vy = uy – gt = u sinθ – gt
  5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
    sx = (u cosθ)t …………..(1)
    sy = (u sinθ)t – \(\frac{1}{2}\) gt2 ………………. (2)
  6. As the projectile starts from x = O, we can use
    sx = x and sy = y.
    Substituting sx = x in equation (1),
    x = (u cosθ) t
    ∴ t = \(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\) …………….. (3)
    Substituting, sy in equation (2),
    y = (u sinθ)t – \(\frac{1}{2}\) gt2 …………… (4)
    Substituting equation (3) in equation (4), we have,
    y = u sin θ (\(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\)) – \(\frac{1}{2}\) (\(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\))2 g
    ∴ y = x (tan θ) – (\(\frac{g}{2 u^{2} \cos ^{2} \theta}\))x2 ………………. (5)
    Equation (5) represents the path of the projectile.
  7. If we put tan θ = A and g/2u2cos2θ = B then equation (5) can be written as y = Ax – Bx2 where A and B are constants. This is equation of parabola. Hence, path of projectile is a parabola.

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 9.
What is a conical pendulum? Show that its time period is given by 2π \(\sqrt{\frac{l \cos \theta}{g}}\), where l is the length of the string, θ is the angle that the string makes with the vertical and g is the acceleration due to gravity.
Answer:
A simple pendulum, Ch i given such a motion that the bob describes a horizontal circle and the string making a constant angle with the vertical describes a cone, is called a conical pendulum.
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 17

  1. Consider a bob of mass m tied to one end of a string of length ‘P and other end is fixed to rigid support.
  2. Let the bob be displaced from its mean position and whirled around a horizontal circle of radius ‘r’ with constant angular velocity ω, then the bob performs U.C.M.
  3. During the motion, string is inclined to the vertical at an angle θ as shown in the figure above.
  4. In the displaced position, there are two forces acting on the bob.
    • The weight mg acting vertically downwards.
    • The tension T acting upward along the string.
  5. The tension (T) acting in the string can be resolved into two components:
    • T cosθ acting vertically upwards.
    • T sinθ acting horizontally towards centre of the circle.
  6. Since, there is no net force, vertical component T cosθ balances the weight and horizontal component T sinθ provides the necessary centripetal force.
    ∴ T cos θ = mg ……………. (1)
    T sin θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mrω2 ……….. (2)
  7. Dividing equation (2) by (1),
    tan θ = \(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) ……….. (3)
    Therefore, the angle made by the string with the vertical is θ = tan-1 (\(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\))
  8. Since we know v = \(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)
    Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 18
    where l is length of the pendulum and h is the vertical distance of the horizontal circle from the fixed point O.

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 10.
Define angular velocity. Show that the centripetal force on a particle undergoing uniform circular motion is -mω2\(\vec{r}\).
Answer:
Angular velocity of a particle is the rate of change of angular displacement.

Expression for centripetal force on a particle undergoing uniform circular motion:
i) Suppose a particle is performing U.C.M in anticlockwise direction.
The co-ordinate axes are chosen as shown in the figure.
Let,
A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
\(\overrightarrow{\mathrm{r}}\) = instantaneous position vector at time t

ii) From the figure,
\(\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}} \mathrm{x}+\hat{\mathrm{j}} \mathrm{y}\)
where, \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) are unit vectors along X-axis and Y-axis respectively.
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 15
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 16
Negative sign shows that direction of acceleration is opposite to the direction of position vector. Equation (3) is the centripetal acceleration.
vii) Magnitude of centripetal acceleration is given by a = ω2r

viii) The force providing this acceleration should also be along the same direction, hence centripetal.
∴ \(\overrightarrow{\mathrm{F}}\) = m\(\overrightarrow{\mathrm{a}}\) = -mω2\(\overrightarrow{\mathrm{r}}\)
This is the expression for the centripetal force on a particle undergoing uniform circular motion.

ix) Magnitude of F = mω2r = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mωv

[Note: The definition of angular velocity is not mentioned in this chapter but is in Ch.2 Mathematical Methods.]

3. Solve the following problems.

Question 1.
An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 sec. What is the velocity of the aeroplane at the take off ?
Answer:
Given: Length of runway (s) = 500 m, t = 30 s
To find: Velocity (y)
Formulae. i) s = ut + \(\frac{1}{2}\) at2
ii) v = u + at
Calculation: As the plane was initially at rest, u = 0
From formula (1),
500 = 0 + \(\frac{1}{2}\) × a × (30)2
∴ 500 = 450 a
∴ a = \(\frac{10}{9}\) m/s2
From formula (ii),
v = 0 + \(\frac{10}{9}\) × 30
∴ v = \(\frac{100}{3}\) m/s = (\(\frac{100}{3} \times \frac{18}{5}\)) km/hr
∴ v = 120 km/hr
The velocity of the aeroplane at the take off is 120 km/hr.

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 2.
A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes. The car covers a distance of 100 m before it stops. Calculate
(i) the average retardation of the car
(ii) time taken by the car to come to rest.
Answer:
Given: u = 120 kmh-1 = 120 × \(\frac{5}{18}\) = \(\frac{100}{3}\) ms-1
s = 100 m, v = 0
To find: i) Average retardation of the car (a)
ii) Time taken by car (t)

Formulae: i) v2 – u2 = 2as
ii) v = u + at
Calculation: From formula (i),
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 5

i) Average retardation of the car is \(\frac{50}{9}\) ms2 (in magnitude).
ii) Time taken by the car to come to rest is 6 s.

Question 3.
A car travels at a speed of 50 km/hr for 30 minutes, at 30 km/hr for next 15 minutes and then 70 km/hr for next 45 minutes. What is the average speed of the car?
Answer:
Given: v1 = 50 km/hr. t1 = 30 minutes = 0.5 hr,
v2 = 30 km/hr, t2 = 15 minutes = 0.25 hr,
v3 = 70 km/hr, t3 = 45 minutes 0.75 hr
To find: Average speed of car (vav)
Formula vav = \(\frac{\text { total path length }}{\text { total time interval }}\)
Calculation:
Path length,
x1 = v1 × t1 = 50 × 0.5 = 25km
x2 = v2 × t2 = 30 × 0.25 = 7.5 km
x3 = v3 × t3 = 70 × 0.75 = 52.5 km
From formula,
vav = \(\frac{x_{1}+x_{2}+x_{3}}{t_{1}+t_{2}+t_{3}}\)
∴ vav = \(\frac{25+7.5+52.5}{0.5+0.25+0.75}=\frac{85}{1.5}\)
∴ vav = 56.66 km/hr

Question 4.
A velocity-time graph is shown in the adjoining figure.
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 1
Determine:

  1. initial speed of the car
  2. maximum speed attained by the car
  3. part of the graph showing zero acceleration
  4. part of the graph showing constant retardation
  5. distance travelled by the car in first 6 sec.

Answer:

  1. Initial speed is at origin i.e. 0 m/s.
  2. Maximum speed attained by car, vmax = speed from A to B = 20 m/s.
  3. The part of the graph which shows zero acceleration is between t = 3 s and t = 6 s i.e., AB. This is because, during AB there is no change in velocity.
  4. The graph shows constant retardation from t = 6 s to t = 8 s i.e., BC.
  5. Distance travelled by car in first 6 s
    = Area of OABDO
    = A(△OAE) + A(rect. ABDE)
    = \(\frac{1}{2}\) × 3 × 20 + 3 × 20
    = 30 + 60
    ∴ Distance travelled by car in first 6 s = 90 m

Question 5.
A man throws a ball to maximum horizontal distance of 80 meters. Calculate the maximum height reached.
Answer:
Given: R = 80m
To find: Maximum height reached (Hmax)
Formula: Rmax = 4Hmax
Calculation: From formula,
∴ Hmax = \(\frac{\mathrm{R}_{\max }}{4}=\frac{80}{4}\) = 20 m
The maximum height reached by the ball is 20m.

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 6.
A particle is projected with speed v0 at angle θ to the horizontal on an inclined surface making an angle Φ (Φ < θ) to the horizontal. Find the range of the projectile along the inclined surface.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 12
i) The equation of trajectory of projectile is given by,
y(tan θ)x – (\(\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\))x2 …………..(1)

ii) In this case to find R substitute,
y = R sinΦ ………….. (2)
x = R cosΦ ………….. (3)

iii) From equations (1), (2) and (3),
we have,
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 13

Question 7.
A metro train runs from station A to B to C. It takes 4 minutes in travelling from station A to station B. The train halts at station B for 20 s. Then it starts from station B and reaches station C in the next 3 minutes. At the start, the train accelerates for 10 sec to reach the constant speed of 72 km/hr. The train moving at the constant speed is brought to rest in 10 sec. at next station.
(i) Plot the velocity-time graph for the train travelling from station A to B to C.
(ii) Calculate the distance between stations A, B, and C.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 4
The metro train travels from station A to station B in 4 minutes = 240 s.
The trains halts at station B for 20 s.
The train travels from station B’ to station C in 3 minutes= 180 s.
∴ Total time taken by the metro train in travelling from station A to B to C
= 240 + 20 + 180 = 440 s.
At start, the train accelerates for 10 seconds to reach a constant speed of 72 km/hr = 20 m/s.
The train moving is brought to rest in 10 s at next station.
The velocity-time graph for the train travelling from station A to B to C is as follows:
Distance travelled by the train from station A to station B
= Area of PQRS
= A ( △PQQ’) A (☐QRR’) + A(SRR’)
= (\(\frac{1}{2}\) × 10 × 20 + (220 × 20) + (\(\frac{1}{2}\) 10 × 20)
= 100 + 4400 + 100
= 4600m = 4.6km

Distance travelled by the train from station B’ to station C
= Area of EFGD
= A(△EFF’) + A(☐F’FGG’) + A(△DGG’)
= (\(\frac{1}{2}\) × 10 × 20) × (160 × 20) + (\(\frac{1}{2}\) × 10 × 20)
= 100 + 3200 + 100
= 3400m = 3.4km

Question 8.
A train is moving eastward at 10 m/sec. A waiter is walking eastward at 1.2m/sec; and a fly is flying toward the north across the waiter’s tray at 2 m/s. What is the velocity of the fly relative to Earth.
Answer:
Given
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 11

Question 9.
A car moves in a circle at the constant speed of 50 m/s and completes one revolution in 40 s. Determine the magnitude of the acceleration of the car.
Answer:
Given: v = 50 m/s, t = 40 s, s = 2πr
To find: acceleration (a)
Formulae: i) v = \(\frac{\mathrm{s}}{\mathrm{t}}\)
ii) a = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
Calculation: From formula (i),
50 = \(\frac{2 \pi \mathrm{r}}{40}\)
∴ r = \(\frac{50 \times 40}{2 \pi}\)
∴ r = \(\frac{1000}{\pi}\) cm
From formula (ii),
a = \(\frac{v^{2}}{r}=\frac{50^{2}}{1000 / \pi}\)
∴ a = \(\frac{5 \pi}{2}\) = 7.85 m/s2
The magnitude of acceleration of the car is 7.85 m/s.

Alternate method:
Given: v = 50 m/s, t = 40 s,
To find: acceleration (a)
Formula: a = rω2 = vω
Calculation: From formula,
a = vω
= v(\(\frac{2 \pi}{\mathrm{t}}\))
= 50(\(\frac{2 \times 3.142}{40}\))
= \(\frac{5}{2}\) × 3.142
∴ a = 7.85m/s2

Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane

Question 10.
A particle moves in a circle with constant speed of 15 m/s. The radius of the circle is 2 m. Determine the centripetal acceleration of the particle.
Answer:
Given: v = 15 m/s, r = 2m
To find: Centripetal acceleration (a)
Formula: a = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
Calculation: From formula,
a = \(\frac{(15)^{2}}{2}=\frac{225}{2}\)
∴ a = 112.5m/s2
The centripetal acceleration of the particle is 112.5 m/s2.

Question 11.
A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50 m? Assume g = 10 m s-2.
Answer:
Given: 40 ≤ R ≤ 50, θ = 300, g = 10 m/s2
To find: Range of initial velocity (u)
Formula: R = \(\frac{\mathrm{u}^{2} \sin (2 \theta)}{\mathrm{g}}\)
Calculation: From formula,
The range of initial velocity,
Maharashtra Board Class 11 Physics Solutions Chapter 3 Motion in a Plane 14
∴ 21.49m/s ≤ u ≤ 24.03m/s
The range of initial velocity should be between 21.49 m/s ≤ u ≤ 24.03 m/s.

11th Physics Digest Chapter 3 Motion in a Plane Intext Questions and Answers

Can you recall? (Textbook Page No. 30)

Question 1.
What ¡s meant by motion?
Answer:
The change ¡n the position of an object with respect to its surroundings is called motion.

Question 2.
What Is rectilinear motion?
Answer:
Motion in which an object travels along a straight line is called rectilinear motion.

Question 3.
What is the difference between displacement and distance travelled?
Answer:

  • Displacement is the shortest distance between the initial and final points of movement.
  • Distance is the actual path followed by a body between the points in which it moves.

Question 4.
What is the difference between uniform and non-uniform motion?
Answer:

  • A body is said to have uniform motion if it covers equal distances in equal intervals of time.
  • A body is said to have non-uniform motion if it covers unequal distances in equal intervals of time.

Internet my friend (Textbook Page No. 44)

i. hyperphysics.phy-astr.gsu.eduJhbase/mot.html#motcon
ii. www .college-physics.comlbook/mechanics
[Students are expected to visit the above mentioned webs ires and collect more information.]

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

1. Choose the most correct option.

Question A.
A sample of pure water, whatever the source always contains …………. by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
Answer:
a. 88.9

Question B.
Which of the following compounds can NOT demonstrate the law of multiple proportions?
a. NO, NO2
b. CO, CO2
c. H2O, H2O2
d. Na2S, NaF
Answer:
d. Na2S, NaF

Question C.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
a. – 40°
b. + 40°
c. – 80°
d. – 20°
Answer:
a. – 40°

Question D.
SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
Answer:
b. Ampere

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question E.
In the reaction N2 + 3H2 → 2NH3, the ratio by volume of N2, H2 and NH3 is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
Answer:
d. gaseous volumes

Question F.
Which of the following has maximum number of molecules ?
a. 7 g N2
b. 2 g H2
c. 8 g O2
d. 20 g NO2
Answer:
b. 2 g H2

Question G.
How many g of H2O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
Answer:
a. 4.5

Question H.
The number of molecules in 22.4 cm3 of nitrogen gas at STP is
a. 6.022 × 1020
b. 6.022 × 1023
c. 22.4 × 1020
d. 22.4 × 1023
Answer:
a. 6.022 × 1020

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question I.
Which of the following has the largest number of atoms ?
a. 1g Au(s)
b. 1g Na(s)
c. 1g Li(s)
d. 1g Cl2(g)
Answer:
c. 1g Li(s)

2. Answer the following questions.

Question A.
State and explain Avogadro’s law.
Answer:
i. In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.

ii. Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 1
According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 2

Question B.
Point out the difference between 12 g of carbon and 12 u of carbon.
Answer:
12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom.

Question C.
How many grams does an atom of hydrogen weigh ?
Answer:
The mass of a hydrogen atom is 1.6736 × 10-24 g.

Question D.
Calculate the molecular mass of the following in u.
a. NH3
b. CH3COOH
c. C2H5OH
Answer:
i. Molecular mass of NH3 = (1 × Average atomic mass of N) + (3 × Average atomic mass of H)
= (1 × 14.0 u) +(3 × 1.0 u)
= 17 u

ii. Molecular mass of CH3COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u)
= 60 u

iii. Molecular mass of C2H5OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u)
= 46 u
Ans: i. The molecular mass of NH3 = 17 u
ii. The molecular mass of CH3COOH = 60 u
iii. The molecular mass of C2H5OH = 46 u

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question E.
How many particles are present in 1 mole of a substance ?
Answer:
The number of particles in one mole is 6.0221367 × 1023.

Question F.
What is the SI unit of amount of a substance ?
Answer:
The SI unit for the amount of a substance is mole (mol).

Question G.
What is meant by molar volume of a gas ?
Answer:
The volume occupied by one mole of a gas at standard temperature (0 °C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm3.

Question H.
State and explain the law of conservation of mass.
Answer:
Law of conservation of mass:

  • The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
  • Antoine Lavoisier who is often referred to as the father of modem chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
  • Both his experiments resulted in increased weight of products.
  • After several experiments, in burning of phosphorus, he found that the weight gained by the phosphoms was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
  • When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.

Question I.
State the law of multiple proportions.
Answer:
The law states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

3. Give one example of each

Question A.
Homogeneous mixture
Answer:
Homogeneous mixture: Solution (An aqueous solution of sugar)

Question B.
Heterogeneous mixture
Answer:
Heterogeneous mixture: Suspension (of sand in water)

Question C.
Element
Answer:
Element: Gold

Question D.
Compound
Answer:
Compound: Distilled water.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

4. Solve problems :

Question A.
What is the ratio of molecules in 1 mole of NH3 and 1 mole of HNO3.
Answer:
One mole of any substance contains particles equal to 6.022 × 1023.
1 mole of NH3 = 6.022 × 1023 molecules of NH3
I mole of HNO3 = 6.022 × 1023 molecules of HNO3
∴ Ratio = \(\frac{6.022 \times 10^{23}}{6.022 \times 10^{23}}\) = 1 : 1
Ans: The ratio of molecules is = 1 : 1.

Question B.
Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 3
Molar volume of a gas = 22.4 dm3 mol-1 = 22.4 L at STP
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 4
Ans: Number of moles of hydrogen = 0.02 mol

Question C.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
∴ Mass of 18 atoms of hydrogen = 18 × 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u

Question D.
Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ x = \(\frac{254 \mathrm{u}}{127 \mathrm{u}}\) = 2 atoms

b. 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol-1
Now,
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 5
Now,
Number of atoms = Number of moles × Avogadro’s constant
= 2 mol × 6.022 × 1023 atoms/mol
= 12.044 × 1023 atoms
= 1.2044 × 1024 atoms
Ans. i.Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = 1.2044 × 1024 atoms

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question E.
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. 5 mg carbon = 5 × 10-3 g carbon
Atomic mass of carbon = 12 u
∴ Molar mass of carbon 12 g mol-1
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 6

b. 12 mg carbon = 12 × 10-3 g carbon
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 7
Number of atoms = Number of moles × Avogadro’s constant
Number of atoms of carbon = 1 × 10-3 mol × 6.022 × 1023 atoms/mol
= 6.022 × 1020 atoms
Ans: Number of moles of carbon in his homework writing = 4.167 × 10-4 mol
Number of atoms of carbon in 12 mg homework writing = 6.022 × 1020 atoms

Question F.
Arjun purchased 250 g of glucose (C6H12O6) for Rs 40. Find the cost of glucose per mole.
Answer:
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C6H12O6
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C6H12O6).
Molecular mass of glucose
= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)
=180 u
∴ Molar mass of glucose = 180 g mol-1
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 8
Now,
\(\frac {250}{180}\) mol of glucose cost = Rs 40
1 mol glucose cost = x
∴ x = \(\frac{40 \times 180}{250}\) = Rs 28.8/mol of glucose
Ans. The cost of glucose per mole is Rs 28.8.

[ Calculation using log table:
\(\frac{40 \times 180}{250}\)
= Antilog10 [log10(40) + log10(180) + log10(250)]
= Antilog10 [1.6021 + 2.2553 – 2.3979]
= Antilog10 [1.4595] = 28.80 ]

Question G.
The natural isotopic abundance of 10B is 19.60% and 11B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron(B)
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 9
Ans. Average atomic mass of boron = 10.84 u

Question H.
Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 °C
b. 30 °C
Answer:
a. Given: Temperature in degree Celsius =40°C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 40 °C in the formula,
°F = \(\frac {9}{5}\) (°C)+32
= \(\frac {9}{5}\) (40) + 32
= 72 + 32
= 104 °F

b. Given: Temperature in degree Celsius = 30 °C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 30 °C in the formula,
°F = \(\frac {9}{5}\)(°C) + 32
= \(\frac {9}{5}\)(30) + 32
= 54 + 32
= 86 °F
Ans: i. The temperature 40 °C corresponds to 104 °F.
ii. The temperature 30 °C corresponds to 86 °F.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question I.
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Calculator: Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH3COOH
= (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12 u) + (4 × 1 u) + (2 × 16 u) = 60 u
∴ Molar mass of acetic acid = 60 g mol-1
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 10
Now,
Number of molecules of acetic acid = Number of moles × Avogadro’s constant
= 0.367 mol × 6.022 × 1023 molecules/mol
= 2.210 × 1023 molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 × 1023 molecules

Question J.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 11
Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 32 = 64 g) of oxygen to give (2 × 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g

Question K.
Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles × Avogadro’s constant
= 0.4 mol × 6.022 × 1023 atoms/mol
= 2.4088 × 1023 atoms of N

b. 1.6 g of Sulphur (S)
Molar mass of sulphur = 32 g mol-1
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 12
Number of atoms of S = Number of moles × Avogadro’s constant
= 0.05 mol × 6.022 × 1023 atoms/mol
= 0.3011 × 1023 atoms
= 3.011 × 1022 atoms of S
Ans: a. Number of nitrogen atoms in 0.4 mole = 2.4088 × 1023 atoms of N
b. Number of sulphur atoms in 1.6 g = 3.011 × 1022 atoms of S

Question L.
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Answer:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
% of oxygen = \(\frac{1.2}{3.2}\) × 100 = 37.5%
% of metal = \(\frac{2.0}{3.2}\) × 100 = 62.5%
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g
% of oxygen = \(\frac{0.85}{2.27}\) × 100 = 37.44 ≈ 37.5%
% of metal = \(\frac{1.42}{2.27}\) × 100 = 62.56 ≈ 62.5%
Therefore, irrespective of the source, the given compound contains same elements in the same proportion. The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight”. Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question M.
In two moles of acetaldehyde (CH3CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer:
Molecular formula of acetaldehyde: C2H4O
Moles of acetaldehyde = 2 mol
a. Number of moles of carbon atoms = Moles of acetaldehyde × Number of carbon atoms
= 2 × 2
= 4 moles of carbon atoms

b. Number of moles of hydrogen atoms = Moles of acetaldehyde × Number of hydrogen atoms
= 2 × 4
= 8 moles of hydrogen atoms

c. Number of moles of oxygen atoms = Moles of acetaldehyde × Number of oxygen atoms
= 2 × 1
= 2 moles of oxygen atoms

d. Number of molecules of acetaldehyde = Moles of acetaldehyde × Avogadro number
= 2 mol × 6.022 × 1023 molecules/mol
= 12.044 × 1023 molecules of acetaldehyde
Ans: i. Number of moles of carbon, hydrogen and oxygen are 4, 8, 2 respectively,
ii. Number of molecules of acetaldehyde = 12.044 × 1023

Question N.
Calculate the number of moles of magnesium oxide, MgO in
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of Mg = 24 and O = 16)
Answer:
Given: i. Mass of MgO = 80 g
ii. Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae: Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
Calculation: i. Molecular mass of MgO = (1 × Average atomic mass of Mg) + (1 × Average atomic mass of O)
= (1 × 24u) + (1 × 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol-1
Mass of MgO = 80 g
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{80 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 2 mol

ii. Mass of MgO = 10 g, Molar mass of MgO = 40 g mol-1
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{10 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.25 mol
Ans: i. The number of moles in 80 g of magnesium oxide, MgO = 2 mol
ii. The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol

Question O.
What is volume of carbon dioxide, CO2 occupying by i. 5 moles and ii. 0.5 mole of CO2 gas measured at STP.
Answer:
Given: i. Number of moles of CO2 = 5 mol
ii. Number of moles of CO2 = 0.5 mol
To find: Volume at STP
Formula: Number of moies of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Calculation: Molar volume of a gas 22.4 dm3 mol-1 at STP.
Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
∴ i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5mol × 22.4 dm3 mol-1 = 112 dm3
ii. Volume of the gas at STP Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm3 mol-1 = 11.2 dm3
Ans: i. Volume of 5 mol of CO2 = 112 dm3
ii. Volume of 0.5 mol of CO2 = 11.2 dm3

Question P.
Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP. Molar mass of KClO3 is 122.5 g mol-1.
Answer:
The molecular formula of potassium chlorate is KClO3.
Required chemical equation:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 13
2 moles of KClO3 = 2 × 122.5 = 245 g
3 moles of O2 at STP occupy = (3 × 22.4 dm3) = 67.2 dm3
Thus, 245 g of potassium chlorate will liberate 67.2 dm3 of oxygen gas.
Let ‘x’ gram of KClO3 liberate 6.72 dm3 of oxygen gas at S.T.P.
∴ x = \(\frac{245 \times 6.72}{67.2}\) = 24.5 g
Ans: Mass of potassium chlorate required = 24.5 g

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question Q.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH2)2CO. Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH2)2CO
Molar mass of urea = 60 g mol-1
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 14
∴ Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 2.247 × 1023 atoms of hydrogen
∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 1.124 × 1023 atoms of nitrogen
∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 0.562 × 1023 atoms of carbon
∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 0.562 × 1023 atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 1023 atoms of H, 1.124 × 1023 atoms of N, 0.562 × 1023 atoms of C and 0.562 × 1023 atoms of O.

Question R.
Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Answer:
Given: Mass of sulphur (reactant) = 16 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 32 g oxygen to form 64 g of sulphur dioxide as follows:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 15
Hence, (0.5 × 32 = 16 g) of sulphur will combine with (0.5 × 32 = 16 g) of oxygen to give (0.5 × 64 = 32 g) sulphur dioxide.
Ans: Mass of sulphur dioxide produced = 32 g

5. Explain

Question A.
The need of the term average atomic mass.
Answer:

  • Several naturally occurring elements exist as a mixture of two or more isotopes.
  • Isotopes have different atomic masses.
  • The atomic mass of such an element is the average of atomic masses of its isotopes.
  • For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.

Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.

Question B.
Molar mass.
Answer:
i. The mass of one mole of a substance (element/compound) in grams is called its molar mass.
ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
e.g.

Element Atomic mass (u) Molar mass (g mol-1)
H 1.0 1 0
C 12.0 12.0
O 16.0 16.0

iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
e.g.

Polyatomic substance Molecular/formula mass (u) Molar mass (g mol-1)
O2 32.0 32.0
H2O 18.0 18.0
NaCl 58.5 58.5

Question C.
Mole concept.
Answer:

  • Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
  • Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon -12 isotope.
  • One mole is the amount of substance which contains 6.0221367 × 1023 particles/entities.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Question D.
Formula mass with an example.
Answer:

  • The formula mass of a substance is the sum of atomic masses of the atoms present in the formula.
  • In substances such as sodium chloride, positive (sodium), and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na+) ion is surrounded by six chlorides (Cl) ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units.
  • Therefore, NaCl is just the formula that is used to represent sodium chloride though it is not a molecule.
  • In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass.

e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

Question E.
Molar volume of gas.
Answer:
i. It is more convenient to measure the volume rather than mass of the gas.
ii. It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm3 at standard temperature (0 °C) and pressure (1 atm) (STP).
iii. The volume of 22.4 dm3 at STP is known as molar volume of a gas.
iv. The relationship between number of moles and molar volume can be expressed as follows:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 16
[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol-1]

Question F.
Types of matter (on the basis of chemical composition).
Answer:
Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.

They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
1. Metals:

  • They have a lustre (a shiny appearance).
  • They conduct heat and electricity.
  • They can be drawn into wire (ductile).
  • They can be hammered into thin sheets (malleable).
  • e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.

2. Nonmetals:

  • They have no lustre, (except diamond, iodine)
  • They are poor conductors of heat and electricity, (except graphite)
  • They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine

3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.

ii. Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.

Mixtures are of two types:

  • Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
    e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
  • Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
    e.g. Suspension, which contains insoluble solid in a liquid.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

11th Chemistry Digest Chapter 1 Some Basic Concepts of Chemistry Intext Questions and Answers

Can you tell? (Textbook Page No. 1)

Question 1.
Which are mixtures and pure substances from the following?
i. Sea water
ii. Gasoline
iii. Skin
iv. A rusty nail
v. A page of textbook
vi. Diamond
Answer:

No. Material Pure substance or mixture
i. Seawater Mixture
ii. Gasoline Mixture
iii. Skin Mixture
iv. A rusty nail Mixture
V. A page of textbook Mixture
vi. Diamond Pure substance

Can you tell? (Textbook Page No. 2)

Question 1.
Classify the following as element and compound.
i. Mercuric oxide
ii. Helium gas
iii. Water
iv. Table salt
v. Iodine
vi. Mercury
vii. Oxygen
viii. Nitrogen
Answer:

No. Material Element or compound
i. Mercuric oxide Compound
ii. Helium gas Element
iii. Water Compound
iv. Table salt Compound
V. Iodine Element
vi. Mercury Element
vii. Oxygen Element
viii. Nitrogen Element

Can you tell? (Textbook Page No. 6)

Question 1.
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 17
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced.

Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Can you recall? (Textbook Page No. 6)

Question 1.
What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:

  • The smallest indivisible particle of an element is called an atom.
  • A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
  • Every atom of an element has definite mass. The order of magnitude of mass of one atom is 10-27 kg.
  • Isotopes are the atoms of the same element having same atomic number but different mass number.

Try this (Textbook Page No. 8)

Question 1.
Find the formula mass of CaSO4, if atomic mass of Ca = 40.1 u, S =32.1 u and O = 16.0 u.
Solution:
Formula mass of CaSO4
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= (40.1) + 32.1 + (4 × 16.0) = 136.2 u
Ans: Formula mass of CaSO4 = 136.2 u

Can you recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items.

Try this (Textbook Page No. 10)

Question 1.
Calculate the volume in dm3 occupied by 60.0 g of ethane at STP.
Solution:
Given: Mass of ethane at STP = 60.0 g
To find: Volume of ethane
Formulae:
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 18
Calculation: Molar volume of a gas = 22.4 dm3 mol-1 at STP
Molecular mass of ethane = 30 g mol-1
Maharashtra Board Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry 19
∴ Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 2 mol × 22.4 dm3 mol-1 = 44.8 dm3
Ans: Volume of ethane = 44.8 dm3

Activity :

Activity 1.
Collect information from various scientists and prepare charts of their contributions to chemistry.
Answer:

Scientists Contributions
Joseph Louis Gay-Lussac (1778 – 1850) (French chemist and physicist) i. Formulated the gas law.
ii. Collected samples of air at different heights and recorded temperatures and moisture contents.
iii. Discovered that the composition of the atmosphere does not change with increasing altitude.
Amedeo Avogadro (1776 – 1856) (Italian scholar) i. Published article in French journal on determining the relative masses of elementary particles of bodies and proportions by which they enter combinations.
ii. Published a research paper titled “New considerations on the theory of proportions and on the determination of the masses of atoms.”

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 14 Semiconductors Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 14 Semiconductors

1. Choose the correct option.

Question 1.
Electric conduction through a semiconductor is due to:
(A) Electrons
(B) holes
(C) none of these
(D) both electrons and holes
Answer:
(D) both electrons and holes

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 2.
The energy levels of holes are:
(A) in the valence band
(B) in the conduction band
(C) in the band gap but close to valence band
(D) in the band gap but close to conduction band
Answer:
(C) in the band gap but close to valence band

Question 3.
Current through a reverse biased p-n junction, increases abruptly at:
(A) Breakdown voltage
(B) 0.0 V
(C) 0.3V
(D) 0.7V
Answer:
(A) Breakdown voltage

Question 4.
A reverse biased diode, is equivalent to:
(A) an off switch
(B) an on switch
(C) a low resistance
(D) none of the above
Answer:
(A) an off switch

Question 5.
The potential barrier in p-n diode is due to:
(A) depletion of positive charges near the junction
(B) accumulation of positive charges near the junction
(C) depletion of negative charges near the junction,
(D) accumulation of positive and negative charges near the junction
Answer:
(D) accumulation of positive and negative charges near the junction

2. Answer the following questions.

Question 1.
What is the importance of energy gap in a semiconductor?
Answer:

  1. The gap between the bottom of the conduction band and the top of the valence band is called the energy gap or the band gap.
  2. This band gap is present only in semiconductors and insulators.
  3. Magnitude of the band gap plays a very important role in the electronic properties of a solid.
  4. Band gap in semiconductors is of the order of 1 eV.
  5. If electrons in valence band of a semiconductor are provided with energy more than band gap energy (in the form of thermal energy or electrical energy), then the electrons get excited and occupy energy levels in conduction band. These electrons can easily take part in conduction.

Question 2.
Which element would you use as an impurity to make germanium an n-type semiconductor?
Answer:
Germanium can be made an n-type semiconductor by doping it with pentavalent impurity, like phosphorus (P), arsenic (As) or antimony (Sb).

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 3.
What causes a larger current through a p-n junction diode when forward biased?
Answer:
In case of forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction.

Question 4.
On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer:
For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor.

Question 5.
Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?
Answer:

  1. In a p-type semiconductor, holes are majority charge carriers.
  2. When a p-type semiconductor is connected to terminals of a battery, holes, which are not actual charges, behave like a positive charge and get attracted towards the negative terminal of the battery.
  3. During transportation of hole, there is an indirect movement of electrons.
  4. The drift speed of these electrons is less than that in the n-type semiconductors. Mobility of the holes is also less than that of the electrons.
  5. As, electrical conductivity depends on the mobility of charge carriers, the conductivity of a n-type semiconductor is greater than that of p-type semiconductor even when both of these have same level of doping.

3. Answer in detail.

Question 1.
Explain how solids are classified on the basis of band theory of solids.
Answer:
i. The solids can be classified into conductors, insulators and semiconductors depending on the distribution of electron energies in each atom.

ii. As an outcome of the small distances between atoms, the resulting interaction amongst electrons and the Pauli’s exclusion principle, energy bands are formed in the solids.

iii. In metals, conduction band and valence band overlap. However, in a semiconductor or an insulator, there is gap between the bottom of the conduction band and the top of the valence band. This is called the energy gap or the band gap.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 1

iv. For metals, the valence band and the conduction band overlap and there is no band gap as shown in figure (b). Therefore, electrons can easily gain electrical energy when an external electric field is applied and are easily available for conduction.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 2

v. In case of semiconductors, the band gap is fairly small, of the order of 1 eV or less as shown in figure (c). Hence, with application of external electric field, electrons get excited and occupy energy levels in conduction band. These can take part in conduction easily.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 3

vi. Insulators, on the contrary, have a wide gap between valence band and conduction band of the order of 5 eV (for diamond) as shown in figure (d). Therefore, electrons find it very difficult to gain sufficient energy to occupy energy levels in conduction band.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 4

vii. Thus, an energy band gap plays an important role in classifying solids into conductors, insulators and semiconductors based on band theory of solids.

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 2.
Distinguish between intrinsic semiconductors and extrinsic semiconductors
Answer:

Intrinsic semiconductors Extrinsic semiconductors
1. A pure semiconductor is known as intrinsic semiconductors. The semiconductor, resulting
2. Their conductivity is low Their conductivity is high even at room temperature.
3. Its electrical conductivity is a function of temperature alone. Its electrical conductivity depends upon the temperature as well as on the quantity of impurity atoms doped in the structure.
4. The number density of holes (nh) is same as the number density of free electron (ne) (nh = ne). The number density of free electrons and number density of holes are unequal.

Question 3.
Explain the importance of the depletion region in a p-n junction diode.
Answer:
i. The region across the p-n junction where there are no charges is called the depletion layer or the depletion region.

ii. During diffusion of charge carriers across the junction, electrons migrate from the n-side to the p-side of the junction. At the same time, holes are transported from p-side to n-side of the junction.

iii. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

iv. The potential barrier thus developed, prevents continuous flow of charges across the junction. A state of electrostatic equilibrium is thus reached across the junction.

v. Free charge carriers cannot be present in a region where there is a potential barrier. This creates the depletion region.

vi. In absence of depletion region, all the majority charge carriers from n-region (i.e., electron) will get transferred to the p-region and will get combined with the holes present in that region. This will result in the decreased efficiency of p-n junction.

vii. Hence, formation of depletion layer across the junction is important to limit the number of majority carriers crossing the junction.

Question 4.
Explain the I-V characteristic of a forward biased junction diode.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 5

  1. Figure given below shows the I-V characteristic of a forward biased diode.
  2. When connected in forward bias mode, initially, the current through diode is very low and then there is a sudden rise in the current.
  3. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve.
  4. The corresponding voltage is called the knee voltage. It is about 0.7 V for silicon and 0.3 V for germanium.
  5. A diode effectively becomes a short circuit above this knee point and can conduct a very large current.
  6. To limit current flowing through the diode, resistors are used in series with the diode.
  7. If the current through a diode exceeds the specified value, the diode can heat up due to the Joule’s heating and this may result in its physical damage.

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 5.
Discuss the effect of external voltage on the width of depletion region of a p-n junction.
Answer:

  1. A p-n junction can be connected to an external voltage supply in two possible ways.
  2. A p-n junction is said to be connected in a forward bias when the p-region connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
  3. In forward bias connection, the external voltage effectively opposes the built-in potential of the junction. The width of depletion region is thus reduced.
  4. The second possibility of connecting p-n junction is in reverse biased electric circuit.
  5. In reverse bias connection, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased

11th Physics Digest Chapter 14 Semiconductors Intext Questions and Answers

Internet my friend (Textbookpage no. 256)

i. https://www.electronics-tutorials.ws/diode
ii. https://www.hitachi-hightech.com
iii. https://nptel.ac.in/courses
iv. https://physics.info/semiconductors
v. http://hyperphysics.phy- astr.gsu.edu/hbase/Solids/semcn.html

[Students are expected to visit above mentioned links and collect more information regarding semiconductors.]