Category: Class 12

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 13 AC Circuits Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 13 AC Circuits

1. Choose the correct option.

i) If the RMS current in a 50 Hz AC circuit is 5A, the value of the current \(\frac{1}{300}\) seconds after its value becomes zero is
(A) 5\(\sqrt {2}\) A
(B) 5\(\sqrt{\frac{3}{2}}\) A
(C) \(\frac{5}{6}\) A
(D) \(\frac{5}{\sqrt{2}}\) A
Answer:
(B) 5\(\sqrt{\frac{3}{2}}\) A

ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 \(\sqrt {2}\) sin (1000t). The power factor of the combination
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\frac{1}{\sqrt{3}}\)
(C) 0.5
(D) 0.6
Answer:
(A) \(\frac{1}{\sqrt{2}}\)

iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is

Answer:
(B) \(\frac{1}{2 \pi f(2 \pi f L-R)}\)

iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + \(\frac{\pi}{3}\)) A. the power dissipated in the circuit is
(A) 106W
(B) 150W
(C) 5625W
(D) Zero
Answer:
(C) 5625W

v) In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(A) 0.607
(B) 0.707
(C) 0.808
(D) 1
Answer:
(B) 0.707

2. Answer in brief.

i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?
Answer:
Impedance, Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\), where R is the resistance of the lamp, w is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, \(\) decreases. Hence, Z increases.
Power factor, cos Φ = \(\frac{R}{Z}\)
As Z increases, the power factor decreases.
Now, the average power over one cycle,
P_av = v_rms i_rms cos Φ
= V_rms \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) cos Φ
= \(\frac{V_{\mathrm{rms}}^{2}}{\mathrm{Z}} \cos \phi\)
∴ P_av decreases as Z increases and cos Φ decreases.
As the current through the lamp \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) decreases, the brightness of the lamp will decrease when C is increased.

ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ?
Answer:
For an LR circuit, the impedance,
Z_LR = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\), where X_L is the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
Z_LCR = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\) because in the case of an inductor the current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) rad. The decrease in net reactance decreases the total impedance (Z_LCR < Z_LR).

iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?
Answer:
The reactance of a capacitor is X_C = \(\frac{1}{2 \pi f C}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, X_C is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

iv) What is wattless current ?
Answer:
The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.
In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.
[Note : In this case, the power factor is zero.]

v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency
Answer:
The natural frequency of LC circuit is \(\frac{1}{2 \pi \sqrt{L C}}\) ,
where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is

Question 3.
In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate P₁ / P₂ .
Answer:
For a series LR circuit, power factor,

Question 4.
When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) – cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle,


= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 5.
Prove that an ideal capacitor in an AC circuit does not dissipate power
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) + cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle, P_av


= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 6.
(a) An emf e = e₀ sin ωt applied to a series L – C – R circuit derives a current I = I₀ sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer:
(a) Instantaneous power,
P = ei
= (e₀ sin ωt) [i₀ (sin ωt ± Φ)]
= e₀i₀ sin ωt(sin ωt cos Φ ± cos ωt sin Φ)
= e₀i₀ sin² ωt ± e₀i₀ sin Φ sin ωt cos ωt
Average power over one cycle,


= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).

(b) P_av = e_rms i_rms cos Φ
The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than e_rms i_rms It means there is significant loss of power during transportation.

Question 7.
A device Y is connected across an AC source of emf e = e₀ sinωt. The current through Y is given as i = i₀ sin(ωt + π/2)
a) Identify the device Y and write the expression for its reactance.
b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically
d) Draw the phasor diagram for the device Y.
Answer:
(a) The device Y is a capacitor. Its reactance is X_c = \(\frac{1}{\omega C}\),
where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b)

(c) X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). Thus X_C ∝ \(\frac{1}{f}\), where f is the frequency of AC. Suppose C = \(\left(\frac{1000}{2 \pi}\right)\) pF
For f= 100 Hz, X_C = 1 × 10⁷Ω = 10MΩ;
for f = 200 Hz, X_C = 5 MΩ;
for f = 300 Hz, X_C = \(\frac{10}{3}\) MΩ;
for f = 400 Hz, X_C = 2.5 MΩ
for f = 500 Hz, X_C = 2 MΩ and so on

(d)

The phasor representing the peak emf (e₀) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

Question 8.
Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Answer:
Figure shows an inductor of inductance L, capacitor of capacitance C, resistor of resistance R, key K and source (power supply) of alternating emf (e) connected to form a closed series circuit.

We assume the inductor, capacitor and resistor to be ideal. As these are connected in series, at any instant, they carry the same current i = i₀ sin ωt. The voltage across the resistor, e_R = Ri, is in phase with the current. The voltage across the inductor, e_L = X_Li, leads the current by \(\frac{\pi}{2}\) rad and that across the capacitor, e_C = X_Ci, lags behind the current by \(\frac{\pi}{2}\) rad. This is shown in the phasor diagram.

is the effective resistance of the circuit. It is called the impedance.

Question 9.
Compare resistance and reactance.
Answer:
(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit.
The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

Question 10.
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Answer:
Figure 13.8 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a key K and a pure inductor of inductance L to form a closed circuit.

On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,
e’ = -L\(\frac{d i}{d t}\) ………………. (1)
where e’ is the induced emf and i is the current through the inductor. To maintain the current; e and e’ must be equal in magnitude and opposite in direction.

According to Kirchhoff’s voltage law, as the resistance of the inductor is assumed to be zero, we

where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = –\(\frac{e_{0}}{\omega L}\)cos ωt = – \(\frac{e_{0}}{\omega L}\)sin(\(\frac{\pi}{2}\) – ωt)
∴ i = \(\frac{e_{0}}{\omega L}\) sin(ωt – \(\frac{\pi}{2}\)) ……………. (3)
as sin (-θ) = – sin θ
From Eq. (3), i_peak = i₀ = \(\frac{e_{0}}{\omega L}\)
∴ i = i₀ sin(ωt – \(\frac{\pi}{2}\)) ………………. (4)
Comparison of this equation with e = e₀ sin ωt shows that e leads i by \(\frac{\pi}{2}\) rad, i.e., the voltage is ahead of current by \(\frac{\pi}{2}\) rad in phase.

Question 11.
An AC source generating a voltage e = e₀ sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Answer:
Figure 13.12 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the

capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = \(\frac{q}{C}\) ∴ q = CV. q and V are functions of time, with V = e = e₀ sin ωt. The instantaneous current in the circuit is i = \(\frac{d q}{d t}=\frac{d}{d t}\)(CV) = C \(\frac{d v}{d t}\) = C \(\frac{d}{d t}\) (e₀ sin ωt) = ωC e₀ cos ωt
∴ i = \(\frac{e_{0}}{(1 / \omega C)} \sin \left(\omega t+\frac{\pi}{2}\right)=i_{0} \sin \left(\omega t+\frac{\pi}{2}\right)\)
where i₀ = \(\frac{e_{0}}{(1 / \omega C)}\) is the peak value of the current.

Table gives the values of e and i for different values of cot and Fig shows graphs of e and i versus ωt. i leads e by phase angle of \(\frac{\pi}{2}\) rad.

Question 12.
If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec.after if was zero ?
Answer:
Data : f = 50 Hz, i_rms = 5 A, t = \(\frac{1}{600}\) s
The peak value of the current,
i₀ = i_rms\(\sqrt {2}\) = (5)(1.414) = 7.07 A
= i₀sin (2πft)
= 7.07 sin [2π(5o) (\(\frac{1}{600}\))]
= 7.07 sin (\(\frac{\pi}{6}\)) = (7.07)(0.5)
= 3.535 A
This is the required current.

Question 13.
A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Answer:
Data: Power (V_rms i_rms) = 100 W, V_rms = 220V,
f = 50 Hz
The rms current through the bulb,
i_rms = \(\frac{\text { power }}{V_{\mathrm{rms}}}=\frac{100}{220}\) = 0.4545 A
The resistance of the bulb,
R = \(\frac{V_{\mathrm{rms}}}{i_{\mathrm{rms}}}=\frac{220}{(100 / 220)}\) = (22) (22) = 484 Ω

Question 14.
A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak)
in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.
Answer:
Data : C = 15 µF = 15 × 10-6 F, V_rms = 220V, f = 50 Hz,
The capacitive reactance = \(\frac{1}{2 \pi f C}\)

If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

Question 15.
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142)
Answer:
Data : L = 2H, i₀ = 0.25 A, f = 60 Hz, π = 3.142
ωL = 2πfL = 2(3.142)(60)(2) = 754.1 Ω
The effective potential difference across the inductor = ωLi_rms = ωL \(\frac{i_{0}}{\sqrt{2}}\)
= \(\frac{(754.1)(0.25)}{1.414}\) = 133.3 V

Question 16.
Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
Answer:
Data: e = 220 sin 100 πt, L = (\(\frac{1}{\pi}\))H
Comparing e = 220 sin 100 πt with
e = e₀ sin ωt, we get
ω = 100 π ∴ ωL = (100 π) (\(\frac{1}{\pi}\)) = 100 Ω
∴ The instantaneous current through the circuit
= i = \(\frac{e_{0}}{\omega L}\) sin(100 πt – \(\frac{\pi}{2}\))
= \(\frac{220}{100}\) sin (100 πt – \(\frac{\pi}{2}\)) = 2.2 sin (100 πt – \(\frac{\pi}{2}\)) in ampere [assuming that e is in volt.]
i_rms = \(\frac{i_{0}}{\sqrt{2}}=\frac{2.2}{1.414}\) = 1.556 A is the reading of the AC galvanometer connected in the circuit.

Question 17.
A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Data: C = 25 µF = 25 × 10^-6F, L = 0.10H, R = 25 Ω ,
e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with
e = e₀ sin (2πft), we get,
the frequency of the alternating emf as

cos Φ = \(\frac{R}{Z}=\frac{25}{99.2}\) = 0.2520
∴ The phase angle, Φ = cos^-1(0.2520) = 75.40° = 1.316 rad

Question 18.
A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
Answer:
Data : C = 100 µF = 100 × 10^-6 F = 10^-4 F,
R = 50 Ω, L = 0.5H, f = 50 Hz, V_rms = 110 V
∴ ωL = 2πfL = 2 (3.142)(50)(0.5) = 157.1 Ω

2500 + 15700 = 18200 Ω²
∴ Impedance, Z = \(\sqrt {18200}\) Ω = 134.9 Ω
The rms value of the current in the circuit,
i_rms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{110}{134.9} \mathrm{~A}\)
= 0.8154 A

Question 19.
Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
Answer:
Data : R = 10 Ω, power factor = 0.5, f = 100 Hz
Power factor = \(\frac{1}{2 \pi f C R}\)
∴ 0.5 = \(\frac{1}{2(3.142)(100) C(10)}\)
∴ C = \(\frac{1}{3.142 \times 10^{3}}\)
= \(\frac{10 \times 10^{-4}}{3.142}\)
= 3.182 × 10^-4 F
This is the capacity of the capacitor.

Question 20.
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Answer:
Data : f = 50 Hz, i = \(\frac{i_{0}}{\sqrt{2}}\) ∴ \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
i = i₀ sinωt
∴ sinωt = \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
∴ ωt = \(\frac{\pi}{4}\) rad
∴ 2πft = \(\frac{\pi}{4}\)
∴ t = \(\frac{1}{8 f}=\frac{1}{8(50)}=\frac{1}{400}\)
= \(\frac{1000 \times 10^{-3}}{400}\) = 2.5 × 10^-3 s
This is the required time.

Question 21.
Calculate the value of capacitance in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
Answer:
Data : f_r = 10⁶ Hz, L = 101.4 × 10^-6 H

= \(\frac{10000 \times 10^{-10}}{4(3.142)^{2}(101.4)}\) = 2.497 × 10^-10 F
= 249.7 × 10^-12 F = 249.7 picofarad
This is the value of the capacity.

Question 22.
A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
Answer:
Data: C = 10 µF = 10 × 10^-6F = 10^-5F,
L = 100mH = 100 × 10^-3 H = 10^-1 H, V = 25V
For reference, see the solved example (8) above.
\(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴i² = \(\frac{C}{L} V^{2}=\frac{10^{-5}}{10^{-1}}(25)^{2}\)
∴i = 25 × 10^-2 A = 0.25 A
This is the maximum current in the coil.

Question 23.
A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10^-6 F = 10^-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \(\frac{1}{2}\)CV²
The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li²
Here, \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴ L = C\(\frac{V^{2}}{i^{2}}\)
∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10^-4 × 10²
= 10^-2H
This is the value of the inductance.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 12 Electromagnetic Induction Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 12 Electromagnetic Induction

1. Choose the correct option.

i) A circular coil of 100 turns with a cross-sectional area (A) of 1 m² is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(A) 1 Wb
(B) 100 Wb
(C) 50 Wb
(D) 200 Wb
Answer:
(B) 100 Wb

ii) A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
(A) BLv
(B) BLv²
(C) \(\frac{1}{2}\)Blv
(D) \(\frac{2 B l}{\mathrm{v}}\)
Answer:
(A) BLv

iii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
(A) 20 mH
(B) 30 mH
(C) 10 mH
(D) \(\frac{20}{3}\) mH
Answer:
(A) 20 mH

iv) A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
(A) 0.1 V
(B) 1 V
(C) 10 V
(D) 0.01 V
Answer:
(A) 0.1 V

v) What is the energy required to build up a current of 1 A in an inductor of 20 mH?
(A) 10 mJ
(B) 20 mJ
(C) 20 J
(D) 10 J
Answer:
(A) 10 mJ

2. Answer in brief.

i) What do you mean by electromagnetic induction? State Faraday’s law of induction.
Answer:
The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.

Faraday’s laws of electromagnetic induction :
(1) First law ; Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law : The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
[Note : The phenomenon was discovered in 1830 by Joseph Henry (1797-1878), US physicist, and independently in 1832 by Michael Faraday (1791 -1867), British chemchemist and physicist.]

ii) State and explain Lenz’s law in the light of principle of conservation of energy.
Answer:
Lenz’s law : The direction of the induced current is such as to oppose the change that produces it.

The change that induces a current may be (i) the motion of a conductor in a magnetic field or (ii) the change of the magnetic flux through a stationary circuit.

Explanation : Consider Faraday’s magnet-and- coil experiment. If the bar magnet is moved towards the coil with its N-pole facing the coil, as in Fig., the number of magnetic lines of induction (pointing to the left) through the coil increases. The induced current in the coil sets up a magnetic field of its own pointing to the right (as given by Amperes right-hand rule) to oppose the growing flux due to the magnet. Hence, to move the magnet towards the coil against this repulsive flux of the induced current, we must do work. The work done shows up as electric energy in the coil.

When the magnet is withdrawn, with its N-pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases. The induced current reverses its direction to supplement the decreasing flux with its own, as shown in Fig.. Facing the coil along the magnet, the induced current is in the clockwise sense. The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force. Thus, we see that Lenz’s law is a consequence of the law of conservation of energy.

[Note : The above law was discovered by Heinrich Friedrich Emil Lenz (1804-65), Russian physicist.]

iii) What are eddy currents? State applications of eddy currents.
Answer:
Whenever a conductor or a part of it is moved in a magnetic field “cutting” magnetic field lines, or placed in a changing magnetic field, the free electrons in the bulk of the metal start circulating in closed paths equivalent to current-carrying loops. These loop currents resemble eddies in a fluid stream and are hence called eddy or Foucault currents [after Jean Bernard Leon Foucault (1819-68), French physicist, who first detected them].

Applications :
(1) Dead-beat galvanometer : A pivoted moving-coil galvanometer used for measuring current has the coil wound on a light aluminium frame. The rotation of the metal frame in magnetic field produces eddy currents in the frame which opposes the rotation and the coil is brought to rest quickly. This makes the galvanometer dead-beat.

(2) Electric brakes : When a conducting plate is pushed into a magnetic field, or pulled out, very quickly, the interaction between the eddy currents in the moving conductor and the field retards the motion. This property of eddy currents is used as a method of braking in vehicles.

iv) If the copper disc of a pendulum swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?
Answer:
As the copper disc enters and leaves the magnetic field, the changing magnetic flux through it induces eddy current in the disc. In both cases, Fleming’s right hand rule shows that opposing magnetic force damps the motion. After a few swings, the mechanical energy becomes zero and the motion comes to a stop.

Joule heating due to the eddy current warms up the disc. Thus, the mechanical energy of the pendulum is transformed into thermal energy.

v) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
Answer:
Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by
\(\frac{1}{L_{\text {parallel }}}=\frac{1}{L_{1}}+\frac{1}{L_{2}}\) or L_parallel = \(\frac{L_{1} L_{2}}{L_{1}+L_{2}}\)
Hence, the equivalent inductance is less than the inductance of either coil.

Question 3.
In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity ω about an axis perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field B acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.
Answer:
Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction \(\vec{B}\) (see the figure in the above Note for reference). \(\vec{B}\) points downwards. Let the constant angular speed of the disc be ω.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is
\(\frac{d A}{d t}\) = frequency of rotation x dA = fdA
where f = \(\frac{\omega}{2 \pi}\) is the frequency of rotation.
.’. \(\frac{d A}{d t}=\frac{\omega}{2 \pi}\) (2 πr dr) = ωr dr
The total emf induced between the axle and the rim of the rotating disc is
\(|e|=\int B \frac{d A}{d t}=\int_{0}^{R} B \omega r d r=B \omega \int_{0}^{R} r d r=B \omega \frac{R^{2}}{2}\)
For anticlockwise rotation in \(\vec{B}\) pointing down, the axle is at a higher potential.

Question 4.
A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 × 10^-4 T. What is the value of induced emf in the wire?
Answer:
Data : l = 20 m, v = 10 m/s. B = 5 × 10^-5 T
The magnitude of the induced emf,
|e| = Blv = (5 × 10^-5)(20)(10) = 10^-2V = 10 mV

Question 5.
A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate
(a) The area swept out per second by the radius of the disc,
(b) The flux cut per second by a radius of the disc,
(c) The induced emf in the disc.
Answer:
Data: R = 0.3m, f = 20 rps, B = 0.2T
(a) The area swept out per unit time by a given radius = (the frequency of rotations) × (the area swept out per rotation) = f(πr²)
= (20)(3.142 × 0.09) = 5.656 m²

(b) The time rate at which a given radius cuts magnetic flux
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B f(πr²)
= (0.2)(5.656) = 1.131 Wb/s

(c) The magnitude of the induced emf,
|e| = \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = 1.131 V

Question 6.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?
Answer:
Data: M = 1.5 H, I_1i = 0, I_1f = 10A, ∆f = 0.2s
The flux linked per unit turn with the second coil due to current I₁ in the first coil is
Φ₂₁ = MI₁
Therefore, the change in the flux due to change in I₁ is
∆₂₁ =M(∆I₁) = M(I_1f – I_1i) = 1.5 (10 – 0)
= 15 Wb
[Note: The rate of change of flux linkage is M(∆I₁/∆t) = 15/0.2 = 75 Wb/s] .

Question 7.
A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm2 and 150 turns (N_c) is wound tightly around the centre of the solenoid. If a current of 3.0A flows through the solenoid, calculate :
(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (µ₀ = 4π × 10^-7 H/m)
Answer:
Data: n = 1.5 × 10³ m , A = 25 × 10^-4 m²,
N_c = 150, I = 3A, ∆t = 0.5s,
µ₀ = 4π × 10^-7 H/m
(a) Magnetic flux density inside the solenoid,
B = u₀ nI = (4π × 10^-7)(1500)(3)
= 5.656 × 10^-3 T = 5.656 mT

(b) Flux per unit turn through the coils of the solenoid, Φ_m = BA
Since the coil C is wound tightly over the solenoid, the flux linkage of C is
N_CΦ_m = N_CBA = (150)(5.656 × 10^-3)(25 × 10^-4)
= 2.121 × 10^-3 Wb = 2.121 mWb

(c) Initial flux through coil C,
Φ_i = N_CΦ_m = 2.121 × 10^-3 Wb
Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is Φ_f = -2.121 × 10^-3 Wb
Therefore, the average emf induced in coil C,
e = \(-\frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}=-\frac{(-2.121-2.121) \times 10^{-3}}{0.5}\)
= 2 × 4.242 × 10^-3 = 8.484 × 10^-3 V = 8.484 mV

Question 8.
A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.
Answer:
Data: N = 2000, A_i = 1.5 × 10^-4 m², A_f = 0,
B = 0.6T, ∆t = 0.2s
Initial flux, NΦ_f = NBA_i = 2000(0.6)(1.5 × 10^-4)
= 0.18 Wb
Final flux, NΦ_f = 0, since the coil is withdrawn out of the field.
Induced emf,e = \(-N \frac{\Delta \Phi_{\mathrm{m}}}{\Delta t}=-N \frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}\)
∴ e = \(-\frac{0-0.18}{0.2}\) = 0.9V

Question 9.
An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10^-5 T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.
Answer:
Data : l = 50 m, B = 6 × 10^-5T, v = 400 m/s
The magnitude of the induced emf,
|e| = Blv = (6 × 10^-5)(400)(50) = 1.2V

Question 10.
A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of the wire is f, calculate the amplitude of the alternating emf induced in the wire.
Answer:
In one rotation, the wire traces out a circle of radius R, i.e., an area A = πR².
Therefore, the rate at which the wire traces out the area is
\(\frac{d A}{d t}\) = frequency or rotation × A = fA
If the angle between the uniform magnetic field \(\vec{B}\) and the rotation axis is θ, the magnitude of the induced emf is
|e|= B\(\frac{d A}{d t}\) cosθ = BfA cosθ = Bf(πR²)cosθ
so that the required amplitude is equal to Bf(πR²).

Question 11.
Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long traveling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (B_v) is given to be 5 × 10^-5T.
Answer:
Data: l = 1.75 m, v = 50 km/h = 50 × \(\frac{5}{18}\) m/s.
B_v = 5 × 10^-5 T
The area swept out by the wing per unit time = 1v.
∴ The magnetic flux cut by the wing per unit time
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B_v(lv)
=(5 × 10^-5)(1.75)(50 × \(\frac{5}{18}\))= 121.5 × 10^-5
= 1.215 mWb/s
Therefore, the magnitude of the induced emf,
|e| =1.215 mV
[Note: In the northern hemisphere, the vertical com ponent of the Earth’s magnetic induction is downwards. Using Fleming’s right hand rule, the port (left) wing-tip would be positive.]

Question 12.
The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil.
Answer:
Data: M = 10 mH = 10^-2 H, I_1i = 5A, I_1f = 1 A,
∆t = 0.2s
The mutually induced emf in coil 2 due to the changing current in coil 1,
e₂₁ = \(-M \frac{\Delta I_{1}}{\Delta t}=-M \frac{I_{1 \mathrm{f}}-I_{1 \mathrm{i}}}{\Delta t}\)
= -(10^-2) \(\left(\frac{1-5}{0.2}\right)\) = 0.2 V

Question 13.
An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?
Answer:
Data: |e₂| = 9.6 × 10^-2 V, dI₁/dt = 1.2 A/s
|e₂| = M\(\frac{d I_{1}}{d t}\)
Mutual inductance,
M = \(\frac{\left|e_{2}\right|}{d I_{1} / d t}=\frac{9.6 \times 10^{-2}}{1.2}\)
= 8 × 10^-2 H
= 80 mH

Question 14.
A long solenoid of length l, cross-sectional area A and having N₁ turns (primary coil) has a small coil of N₂ turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.
Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current Is through the solenoid, the uniform magnetic field inside the solenoid is
B = µ₀nI_s ……………….. (1)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
Φ_CS = BA = (µ₀nI_s)(πR²) ……………… (2)
Thus, their mutual inductance is
M = \(\frac{N \Phi_{\mathrm{CS}}}{I_{\mathrm{S}}}\) = µ₀πR²nN ……………. (3)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.
replacing n with N₁/l and N with N₂. M = µ₀A = \(\frac{N_{1} N_{2}}{l}\)
[Note: The answer given in the textbook misses out the factor of 1.] .

Question 15.
The primary and secondary coil of a transformer each have an inductance of 200 × 10^-6H. The mutual inductance (M) between the windings is 4 × 10^-6H. What percentage of the flux from one coil reaches the other?
Answer:
Data: L_P = L_S = 2 × 10^-4 H, M = 4 × 10^-6 H
M = K\(\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}\)
The coupling coefficient is
K = \(\frac{M}{\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}}=\frac{4 \times 10^{-6}}{\sqrt{\left(2 \times 10^{-4}\right)^{2}}}=\frac{4 \times 10^{-6}}{2 \times 10^{-4}}\)
= 2 × 10^-2
Therefore, the percentage of flux of the primary reaching the secondary is
0.02 × 100% = 2%

Question 16.
A toroidal ring, having 100 turns per cm of a thin wire is wound on a nonmagnetic metal rod of length 1 m and diameter 1 cm. If the permeability of bar is equal to that of free space (µ₀), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 1 A. Also determine the self-inductance (L) of the coil.
Answer:
Data : l = 1 m, d = 1 cm, n = 100 cm^-1 = 10⁴ m^-1,
I = 100 A, µ₀ = 4π × 10^-7 H/m
The radius of cross section, r = \(\frac{d}{2}\) = 0.5 cm
= 5 × 10^-3 m
(a) Magnetic field inside the toroid,
B = µ₀nI = (4π × 10^-7)(10⁴)(100)
= 0.4 × 3.142 = 1.257 T

(b) Self inductance of the toroid,
L = µ₀2πRn²A = µ₀n²lA = µ₀n²l(πr²)
= (4π × 10^-7)(10⁴)²(1) [π(5 × 10^-3)²]
= π² × 10^-3 = 9.87 × 10^-3 H = 9.87 mH

Question 17.
A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.
[Hint : Part of Maxwell’s equation, applied to a time varying magnetic flux, leads us to the equation \(\oint \vec{E} \cdot \overline{\mathrm{d} l}=\frac{-d \phi_{m}}{d t}\), where \(\vec{E}\) is the electric field induced when the magnetic flux changes at the rate of \(\frac{d \phi_{m}}{d t}\)]
Answer:
The area of the region, A = πs², remains constant while B = B(f) is a function of time. Therefore, the induced emf,
e = \(-\frac{d \Phi_{\mathrm{m}}}{d t}=-\frac{d}{d t}(B A)=-A \frac{d B(t)}{d t}=-\pi s^{2} \frac{d B(t)}{d t}\)
[Note : Emf and electric field are different physical quantities, whose respective SI units are the volt and the volt per metre. The question has accordingly been corrected.]

12th Physics Digest Chapter 12 Electromagnetic induction Intext Questions and Answers

Do you know (Textbook Page No. 274)

Question 1.
If a wire without any current is kept in a magnetic field, then it experiences no force as shown in figure (a). But when the wire is carrying a current into the plane of the paper in the magnetic field, a force will be exerted on the wire towards the left as shown in the figure (b). The field will be strengthened on the right side of the wire where the lines of force are in the same direction as that of the magnetic field and weakened on the left side where the field lines are in opposite direction to that of the applied magnetic field. For a wire carrying a current out of the plane of the paper, the force will act to the right as shown in figure (c).

Answer:
Force on a current-carrying conductor in a magnetic field, \(\vec{F}=I \vec{L} \times \vec{B}\) (Refer unit 10.5). The field due to acurrent-carrying straight conductor is given by right- hand grip rule. As shown in the figure below, the combined field due to a permanent magnet and a current-carrying conductor force the conductor out of the field. The field is strengthened where the two fields are in the same direction and add constructively while the field is weakened where the two fields are opposite in direction.

Use your brain power (Textbook Page No. 282)

Question 1.
It can be shown that the mutual potential energy of two circuits is W = MI₁I₂. Therefore, the mutual inductance (M) may also be defined as the mutual potential energy (W) of two circuits corresponding to unit current flowing in each circuit.
M = \(\frac{W}{I_{1} I_{2}}\)
M = W[I₁ = I₂ = 1]
Answer:
Mutual inductance of two magnetically linked coils equals the potential energy for unit currents in the coils.
1 H = 1 T∙m²/A (= 1 V∙s/A = 1 Ω ∙ s = 1 J/A²)

Use your brain power (Textbook Page No. 284)

Question 1.
Prove that the inductance of parallel wires of length l in the same circuit is given by L = \(\left(\frac{\mu_{0} l}{\pi}\right)\) ln (d / a), where a is the radius of wire and d is separation between wire axes.
Answer:
If l is the current in each wire, from Ampe’re’s law the magnitude of the magnetic field outside each wire is
B = \(\frac{\mu_{0} I}{2 \pi r}\)
By right hand grip rule, the direction of the magnetic field due to both the wires are in the same direction at the point shown. Hence, by the symmetry of the setup, the total magnetic flux through an area dA = l dr shown is two times that due to one wire.

Do you know (Textbook Page No. 285)

Question 1.
The flux rule is the terminology that Feynman used to refer to the law relating magnetic flux to emf. (RP Feynman, Feynman Lectures on Physics, Vol II)
Answer:
Modern applications of Faraday’s law of induction :

• Electric generators and motors
• Dynamos in vehicles
• Transformers
• Induction furnaces (industrial), induction cooking stoves (domestic)
• Radio communication
• Magnetic flow meters and energy meters
• Metal detectors at security checks
• Magnetic hard disk and tape, storage and retrieval
• Graphics tablets
• ATM Credit/debit cards, ATM and point-of-sale (POS) machines
• Pacemakers

Faraday’s second law of electromagnetic induction is referred by some as the “flux rule”.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 9 Current Electricity Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 9 Current Electricity

1. Choose the correct option.

i) Kirchhoff’s first law, i.e., ΣI = 0 at a junction, deals with the conservation of
(A) charge
(B) energy
(C) momentum
(D) mass
Answer:
(A) charge

ii) When the balance point is obtained in the potentiometer, a current is drawn from
(A) both the cells and auxiliary battery
(B) cell only
(C) auxiliary battery only
(D) neither cell nor auxiliary battery
Answer:
(D) neither cell nor auxiliary battery

iii) In the following circuit diagram, an infinite series of resistances is shown. Equivalent resistance between points A and B is

(A) infinite
(B) zero
(C) 2 Ω
(D) 1.5 Ω
Answer:
(C) 2 Ω

iv) Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across 15 Ω resistor to balance the bridge
(A) 10 Ω
(B) 15 Ω
(C) 20 Ω
(D) 30 Ω
Answer:
(D) 30 Ω

v) A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one quarter of the circumference apart are connected to a 24 V battery, having an internal resistance of 0.5 Ω. What is the current flowing through the battery.
(A) 0.5 A
(B) 1A
(C) 2A
(D) 3A
Answer:
(D) 3A

vi) To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a metre bridge. A resistance of 4 Ω is introduced in the right gap. What is the resistance of the bangle if the null point is at 20 cm from the left end?
(A) 2 Ω
(B) 4 Ω
(C) 8 Ω
(D) 16 Ω
Answer:
(A) 2 Ω

2. Answer in brief.

i) Define or describe a Potentiometer.
Answer:
The potentiometer is a device used for accurate measurement of potential difference as well as the emf of a cell. It does not draw any current from the circuit at the null point. Thus, it acts as an ideal voltmeter and it can be used to determine the internal resistance of a cell. It consist of a long uniform wire AB of length L, stretched on a wooden board. A cell of stable emf (E), with a plug key K in series, is connected across AB as shown in the following figure.

ii) Define Potential Gradient.
Answer:
Potential gradient is defined as the potential difference (the fall of potential from the high potential end) per unit length of the wire.

iii) Why should not the jockey be slided along the potentiometer wire?
Answer:
Sliding the jockey on the potentiometer wire decreases the cross sectional area of the wire and thereby affects the fall of potential along the wire. This affects the potentiometer readings. Flence, the jockey should not be slided along the potentiometer wire.

iv) Are Kirchhoff’s laws applicable for both AC and DC currents?
Answer:
Kirchhoff’s laws are applicable to both AC and DC ’ circuits (networks). For AC circuits with different loads, (e.g. a combination of a resistor and a capacitor, the instantaneous values for current and voltage are considered for addition.

[Note : Gustav Robert Kirchhoff (1824-87), German physicist, devised laws of electrical network and, with Robert Wilhelm Bunsen, (1811-99), German chemist, did pioneering work in chemical spectroscopy. He also con-tributed to radiation.]

v) In a Wheatstone’s meter-bridge experiment, the null point is obtained in middle one third portion of wire. Why is it recommended?
Answer:

1. The value of unknown resistance X, may not be accurate due to non-uniformity of the bridge wire and development of contact resistance at the ends of the wire.
2. To minimise these errors, the value of R is so adjusted that the null point is obtained in the middle one-third of the wire (between 34 cm and 66 cm) so that the percentage errors in the measurement of I_X and I_R are minimum and nearly the same.

vi) State any two sources of errors in meterbridge experiment. Explain how they can be minimized.
Answer:
The chief sources of error in the metre bridge experiment are as follows :

1. The bridge wire may not be uniform in cross section. Then the wire will not have a uniform resistance per unit length and hence its resistance will not be proportional to its length.
2. End resistances at the two ends of the wire may be introduced due to
   1. the resistance of the metal strips
   2. the contact resistance of the bridge wire with the metal strips
   3. unmeasured lengths of the wire at the ends because the contact points of the wire with the metal strips do not coincide with the two ends of the metre scale attached.

Such errors are almost unavoidable but can be minimized considerably as follows :

1. Readings must be taken by adjusting the standard known resistance such that the null point is obtained close to the centre of the wire. When several readings are to be taken, the null points should lie in the middle one-third of the wire.
2. The measurements must be repeated with the standard resistance (resistance box) and the unknown resistance interchanged in the gaps of the bridge, obtaining the averages of the two results.

vii) What is potential gradient? How is it measured? Explain.
Answer:
Consider a potentiometer consisting of a long uniform wire AB of length L and resistance R, stretched on a wooden board and connected in series with a cell of stable emf E and internal resistance r and a plug key K as shown in below figure.

Let I be the current flowing through the wire when the circuit is closed.
Current through AB, I = \(\frac{E}{R+r}\)
Potential difference across AB. V_AB = IR
∴ V_AB = \(\frac{E R}{(R+r)}\)
The potential difference (the fall of potential from the high potential end) per unit length of the wire,
\(\frac{V_{\mathrm{AB}}}{L}=\frac{E R}{(R+r) L}\)
As long as E and r remain constant, \(\frac{V_{\mathrm{AB}}}{L}\) will remain constant, \(\frac{V_{\mathrm{AB}}}{L}\) is known as potential gradient along

AB and is denoted by K. Thus the potential gradient is calculated by measuring the potential difference between ends of the potentiometer wire and dividing it by the length of the wire.
Let P be any point on the wire between A and B and AP = l = length of the wire between A and P.
Then V_AP = Kl
∴ V_AP ∝ l as K is constant in a particular case. Thus, the potential difference across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.

viii) On what factors does the potential gradient of the wire depend?
Answer:
The potential gradient depends upon the potential difference between the ends of the wire and the length of the wire.

ix) Why is potentiometer preferred over a voltmeter for measuring emf?
Answer:
A voltmeter should ideally have an infinite resistance so that it does not draw any current from the circuit. However a voltmeter cannot be designed to have infinite resistance. A potentiometer does not draw any current from the circuit at the null point. Therefore, it gives more accurate measurement. Thus, it acts as an ideal voltmeter.

x) State the uses of a potentiometer.
Answer:
The applications (uses) of the potentiometer :

1. Voltage divider :
   The potentiometer can be used as a voltage divider to change the output voltage of a voltage supply.
2. Audio control:
   Sliding potentiometers are commonly used in modern low-power audio systems as audio control devices. Both sliding (faders) and rotary potentiometers (knobs) are regularly used for frequency attenuation, loudness control and for controlling different characteristics of audio signals.
3. Potentiometer as a sensor:
   If the slider of the potentiometer is connected to the moving part of a machine, it can work as a motion sensor. A small displacement of the moving part causes a change in potential which is further amplified using an amplifier circuit. The potential difference is calibrated in terms of displacement of the moving part.
4. To measure the emf (for this, the emf of the standard cell and potential gradient must be known).
5. To compare the emf’s of two cells.
6. To determine the internal resistance of a cell.

xi) What are the disadvantages of a potentiometer?
Answer:
Disadvantages of a potentiometer over a voltmeter :

1. The use of a potentiometer is an indirect measurement method while a voltmeter is a direct reading instrument.
2. A potentiometer is unwieldy while a voltmeter is portable.
3. Unlike a voltmeter, the use of a potentiometer in measuring an unknown emf requires a standard . source of emf and calibration.

xii) Distinguish between a potentiometer and a voltmeter.
Answer:

Potentiometer	Voltmeter
1. A potentiometer is used to determine the emf of a cell, potential difference and internal resistance.	1. A voltmeter can be used to measure the potential difference and terminal voltage of a cell. But it cannot be used to measure the emf of a cell.
2. Its accuracy and sensitivity are very high.	2. Its accuracy and sensitivity are less as compared to a potentiometer.
3. It is not a portable instrument.	3. It is a portable instrument.
4. It does not give a direct reading.	4. It gives a direct reading.

xiii) What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is
(i) increased (ii) decreased.
Answer:
(1) On increasing the current through the potentiometer wire, the potential gradient along the wire will increase. Hence, the position of zero deflection will occur at a shorter length.

(2) On decreasing the current through the potentiometer wire, the potential gradient along the wire will decrease. Hence, the position of zero deflection will occur at a longer length.
[Note : In the usual notation,
E₁ = (\(\frac{I R}{L}\)) l₁ = constant
Hence, (i) E, decreases when I is increased (ii) l₁ increases when I is decreased.]

Question 3.
Obtain the balancing condition in case of a Wheatstone’s network.
Answer:
Wheatstone’s network or bridge is a circuit for indirect measurement of resistance by null com-parison method by comparing it with a standard known resistance. It consists of four resistors with resistances P, Q, R and S arranged in the form of a quadrilateral ABCD. A cell (E) with a plug key (K) in series is connected across one diagonal AC and a galvanometer (G) across the other diagonal BD as shown in the following figure.

With the key K dosed, currents pass through the resistors and the galvanometer. One or more of the resistances is adjusted until no deflection in the galvanometer can be detected. The bridge is then said to be balanced.

Let I be the current drawn from the cell. At junction A, it divides into a current I₁ through P and a current I₂ through S.
I = I₁ + I₂ (by Kirchhoff’s first law).
At junction B, current I_g flows through the galvanometer and current I₁ – I_g flows through Q. At junction D, I₂ and I_g combine. Hence, current I₂ + I_g flows through R from D to C. At junction C, I₁ – I_g and I₂ + I_g combine. Hence, current I₁ + I₂(= I) leaves junction C.

Applying Kirchhoff’s voltage law to loop ABDA in a clockwise sense, we get,
– I₁P – I_gG + I₂S = 0 …………… (1)
where G is the resistance of the galvanometer.
Applying Kirchhoff’s voltage law to loop BCDB in a clockwise sense, we get,
– (I₁ – I_g)Q + (I₂ + I_g)R + I_gG = 0 ………….. (2)
When I_g = 0, the bridge (network) is said to be balanced. In that case, from Eqs. (1) and (2), we get,
I₁P = I₂S …………… (3)
and I₁Q = I₂R ………….. (4)
From Eqs. (3) and (4), we get,
\(\frac{P}{Q}=\frac{S}{R}\)
This is the condition of balance.
Alternative Method: When no current flows through the galvanometer, points B and D must be at the same potential.
∴ V_B = V_D
∴ V_A – V_B = V_A – V_D …………. (1)
and V_B – V_C = V_D – V_C ………… (2)
Now, V_A – V_B = I₁P and V_A – V_D = I₂S ………….. (3)
Also, V_B – V_C = I₁Q and V_D – V_C = I₂R …………. (4)
Substituting Eqs. (3) and (4) in Eqs. (1) and (2), we get,
I₁P = I₂S . ………… (5)
and I₁Q = I₂R …(6)
Dividing Eq. (5) by Eq. (6), we get,
\(\frac{P}{Q}=\frac{S}{R}\)
This is the condition of balance.

[ Note : In the determination of an unknown resistance using Wheatstone’s network, the unknown resistance is connected in one arm of the network (say, AB), and a standard known variable resistance is connected in an adjacent arm. Then, the other two arms are called the ratio arms. Also, because the positions of the cell and galvanometer can be interchanged, without a change in the condition of balance, the branches AC and BD in figure are called the conjugate arms. ]

Question 4.
Explain with neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge.
Answer:
A metre bridge consists of a rectangular wooden board with two L-shaped thick metallic strips fixed along its three edges. A single thick metallic strip separates two L-shaped strips. A wire of length one metre and uniform cross section is stretched on a metre scale fixed on the wooden board. The ends of the wire are fixed to the L-shaped metallic strips.

An unknown resistance X is connected in the left gap and a resistance box R is connected in the right gap as shown in above figure. One end of a centre-zero galvanometer (G) is connected to terminal C and the other end is connected to a pencil jockey (J). A cell (E) of emf E, plug key (K) and rheostat (Rh) are connected in series between points A and B.

Working : Keeping a suitable resistance (R) in the resistance box, key K is closed to pass a current through the circuit. The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows zero deflection. The bridge is then balanced and point D is called the null point and the method is called as null deflection method. The distances l_X and l_R of the null point from the two ends of the wire are measured.

As R, l_X and l_R are known, the unknown resistance X can be calculated.

Question 5.
Describe Kelvin’s method to determine the resistance of a galvanometer by using a meter bridge.
Answer:
Kelvin’s method :
Circuit: The metre bridge circuit for Kelvin’s method of determination of the resistance of a galvanometer is shown in below figure. The galvanometer whose resistance G is to be determined, is connected in one gap of the metre bridge. A resistance box providing a variable known resistance R is connected in the other gap. The junction B of the galvanometer and the resistance box is con-nected directly to a pencil jockey. A cell of emf E, a key (K) and a rheostat (Rh) are connected across AC.

Working : Keeping a suitable resistance R in the resistance box and maximum resistance in the rheostat, key K is closed to pass the current. The rheostat resistance is slowly reduced such that the galvanometer shows about 2 / 3rd of the full-scale deflection.

On tapping the jockey at end-points A and C, the galvanometer deflection should change to opposite sides of the initial deflection. Only then will there be a point D on the wire which is equipotential with point B. The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows no change in deflection. Point D is now called the balance point and Kelvin’s method is thus an equal deflection method. At this balanced condition,
\(\frac{G}{R}=\frac{\text { resistance of the wire of length } l_{G}}{\text { resistance of the wire of length } l_{R}}\)
where I_G = the length of the wire opposite to the galvanometer, I_R = the length of the wire opposite to the resistance box.
If λ = the resistance per unit length of the wire,
\(\frac{G}{R}=\frac{\lambda l_{G}}{\lambda l_{R}}=\frac{l_{G}}{l_{R}}\)
∴ G = R\(\frac{l_{G}}{l_{R}}\)
The quantities on the right hand side are known, so that G can be calculated.
[Note : The method was devised by William Thom-son (Lord Kelvin, 1824-1907), British physicist.]

Question 6.
Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually.
Answer:
A battery of stable emf E is used to set up a potential gradient V/L along a potentiometer wire, where V = potential difference across the length L of the wire. The positive terminals of the cells, whose emf’s (E₁ and E₂) are to be compared, are connected to the high potential terminal A. The negative terminals of the cells are connected to a galvanometer G through a two-way key. The other terminal of the galvanometer is connected to a pencil jockey. The emf E should be greater than both the emf’s E₁ and E₂.

Connecting point P to C, the cell with emf E1 is brought into the circuit. The jockey is tapped along the wire to locate the null point D at a distance l₁ from A. Then,
E₁ = Z₁(V/L)
Now, without changing the potential gradient (i.e., without changing the rheostat setting) point Q (instead of P) is connected to C, bringing the cell with emf E₂ into the circuit. Let its null point D’ be at a distance l₂ from A, so that
E₂ = l₂(V/L)
∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\)
Hence, by measuring the corresponding null lengths l₁ and l₂, E₁/E₂ can be calculated. The experiment is repeated for different potential gradients using the rheostat.
[Note : This method is used when the two emf’s have comparable magnitudes. Then, the errors of measurement of their balancing lengths will also be of comparable magnitudes.]

Question 7.
Describe how a potentiometer is used to compare the emfs of two cells by combination method.
Answer:
A battery of stable emf E is used to set up a potential gradient V/L, along the potentiometer wire, where V = potential difference across length L of the wire. The positive terminal of the cell 1 is connected to the higher potential terminal A of the potentiometer; the negative terminal is connected to the galvanometer G through the reversing key. The other terminal of the galvanometer is connected to a pencil jockey. The cell 2 is connected across the remaining two opposite terminals of the reversing key. The other terminal of the galvanometer is connected to a pencil jockey. The emf E₁ should be greater than the emf E₂; this can be adjusted by trial and error.

Two plugs are inserted in the reversing key in positions 1 – 1. Here, the two cells assist each other so that the net emf is E₁ + E₂. The jockey is tapped along the wire to locate the null point D. If the null point is a distance l₁ from A,
E₁ + E₂ = l₁ (V/L)

For the same potential gradient (without changing the rheostat setting), the plugs are now inserted into position 2 – 2. (instead of 1 – 1). The emf E₂ then opposes E₁ and the net emf is E₁ – E₂. The new null point D’ is, say, a distance l₂ from A and
E₁ – E₂ = l₂ (V/L)
∴ \(\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{l_{1}}{l_{2}}\) ∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}-l_{2}}{l_{1}-l_{2}}\)
Here, the emf E should be greater than E₁ + E₂. The experiment is repeated for different potential gradients using the rheostat.
[Note : This method is used when E₁ » E₂, so that E₁ + E₂ and E₁ – E₂ have comparable magnitudes. Then, the errors of measurement of l₁ and l₂ will also be of comparable magnitudes.]

Question 8.
Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.
Answer:
Principle : A cell of emf £ and internal resistance r, which is connected to an external resistance R, has its terminal potential difference V less than its emf E. If I is the corresponding current,
\(\frac{E}{V}=\frac{I(R+r)}{I R}=\frac{R+r}{R}\) = 1 + \(\frac{r}{R}\) (when R → ∞, V → E)
∴ r = \(\frac{E-V}{V}\) R
Working : A battery of stable emf E’ is used to set up a potential gradient V/L, along the potentiometer wire, where V = potential difference across the length L of the wire. The negative terminal is connected through a centre-zero galvanometer G to a pencil jockey. A resistance box R with a plug key K in series is connected across the cell.

Firstly, key K is kept open; then, effectively, R = ∞. The jockey is tapped on the potentiometer wire to locate the null point D. Let the null length
AD = l₁, so that
E = (V_AB/L)l₁

With the same potential gradient, and a small resistance R in the resistance box, key K is closed. The new null length AD’ = l2 for the terminal potential difference V is found :

R, l₁, and l₂ being known, r can be calculated. The experiment is repeated with different potential gradients using the rheostat or with different values of R.

Question 9.
On what factors does the internal resistance of a cell depend?
Answer:
The internal resistance of a cell depends on :

1. Nature of the electrolyte :
   The greater the conductivity of the electrolyte, the lower is the internal resistance of the cell.
2. Separation between the electrodes :
   The larger the seperation between the electrodes of the cell, the higher is the internal resistance of the cell. This is because the ions have to cover a greater distance before reaching an electrode.
3. Nature of the electrodes.
4. The internal resistance is inversly proportional to the common area of the electrodes dipping in the electrolyte.

Question 10.
A battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with
another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance.
Answer:
Let I₁ and I₂ be the currents through the two branches as shown in below figure. The current through the 2 Q resistance will be (I₁ + I₂) [Kirchhoff’s current law].
Applying Kirchhoff’s voltage law to loop ABCDEFA, we get,
-2(I₁ + I₂) – 1(I₁) + 4 = 0
∴ 3I₁ + 2I₁ = 4 …………… (1)

Applying Kirchhoff’s voltage law to loop BCDEB, we get,
-2(I₁ + I₂) – 1(I₂) + 1 = 0
2I₁ + 3I₂ = 1 ………… (2)
Multiplying Eq. (1) by 2 and Eq. (2) by 3, we get,
6I₁ + 4I₂ = 8 ………….(3)
and 6I₁ + 9I₂ = 3 ……………. (4)
Subtracting Eq. (4) from Eq. (3), we get,
– 5I₂ =5
∴ I₂ = -1A
The minus sign shows that the direction of current I2 is opposite to that assumed. Substituting this value of 12 in Eq. (1), we get,
3I₁ + 2(-1) = 4
∴3I₁ = 4 + 2 = 6
∴ I₁ = 2A
Current through the 2 0 resistance = I₁ + I₂ = 2 – 1
= 1 A. It is in the direction as shown in the figure.
[Note : We may as well consider loop ABEFA and write the corresponding equation. But it does not provide any additional information.]

Question 11.
Two cells of emf 1.5 Volt and 2 Volt having respective internal resistances of 1 Ω and 2 Ω are connected in parallel so as to send current in same direction through an external resistance of 5 Ω. Find the current through the external resistance.
Answer:
Let I₁ and I₂ be the currents through the two branches as shown in below figure. The current through the 5 Q resistor will be I₁ + I₂ [Kirchhoff’s current law].

Applying Kirchhoff’s voltage law to loop w ABCDEFA, we get,
– 5(I₁ + I₂) – I₁ + 1.5 = 0
∴ 6I₁ + 5I₂ = 1.5 ……………. (1)
Applying Kirchhoff’s voltage law to loop BCDEB, we get,
– 5(I₁ + I₂) – 2I₂ + 2 = 0
∴ 5I₁ + 7I₂ = 2 ……………(2)
Multiplying Eq. (1) by 5 and Eq. (2) by 6, we get,
30I₁ + 25I₂ = 7.5 …………… (3)
and 30I₁ + 42I₂ = 12 …………. (4)
Subtracting Eq. (3) from Eq. (4), we get,
17I₂ = 4.5
∴ I₂ = \(\frac{4.5}{17}\) A
Substituting this value of I2 in Eq. (1), we get,
6I₁ + 5(\(\frac{4.5}{17}\)) = 1.5
∴ 6I₁ + \(\frac{22.5}{17}\) = 1.5
∴ 6I₁ = 1.5 – \(\frac{22.5}{17}\) = \(\frac{28.5-22.5}{17}\)
= \(\frac{3}{17}\)
∴ I₁ = \(\frac{3}{17 \times 6}=\frac{0.5}{17}\) A
Current through the 5 Q resistance (external resistance)
= I₁ + I₂ = \(\frac{0.5}{17}+\frac{4.5}{17}=\frac{5}{17}\) A

Question 12.
A voltmeter has a resistance 30 Ω. What will be its reading, when it is connected across a cell of emf 2 V having internal resistance 10 Ω?
Answer:
Data: E = 2V, r = 10 Ω, R = 30 Ω
The voltmeter reading, V = IR
= (\(\frac{E}{R+r}\)) R
= (\(\frac{2}{30+10}\)) 30
= (\(\frac{2}{40}\)) 30
= 1.5 V

Question 13.
A set of three coils having resistances 10 Ω, 12 Ω and 15 Ω are connected in parallel. This combination is connected in series with series combination of three coils of the same resistances. Calculate the total resistance and current through the circuit, if a battery of emf 4.1 Volt is used for drawing current.
Answer:
Below figure shows the electrical network. For resistances 10 Ω, 12 Ω and 15 Ω connected in parallel the equivalent resistance (R_p) is given by,

For resistance R_p, 10 Ω, 12 Ω and 15 Ω connected in series, the equivalent resistance,
R_s = 4 + 10 + 12 + 15 = 41 Ω
Thus, the total resistance = R_s = 41 Ω
Now, V = IR_s
∴ 4.1 = 1 × 41
∴ I = 0.1A
The total resistance and current through the circuit are 41 Ω and 0.1 A respectively.

Question 14.
A potentiometer wire has a length of 1.5 m and resistance of 10 Ω. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Ω. Calculate the potential drop per centimeter of the wire.
Answer:
Data : L = 1.5 m, R = 10 Ω, E = 4 V, r = 5 Ω

The potential drop per centimeter of the wire is 0.0178 \(\frac{\mathrm{V}}{\mathrm{cm}}\)

Question 15.
When two cells of emfs. E₁ and E₂ are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.
Answer:
Data : l₁ = 2.7m (cells assisting),
l₂ = 0.3 m (cells opposing)
E₁ + E₂ = Kl₁ and E₁ – E₂ = Kl₂
∴\(\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{K l_{1}}{K l_{2}}\)
∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}+l_{2}}{l_{1}-l_{2}}=\frac{2.7+0.3}{2.7-0.3}=\frac{3}{2.4}=\frac{30}{24}\) = 1.25
The ratio of the emf’s of the two cells is 1.25.

Question 16.
The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
Answer:
Data :R = 10 Ω, l₁ = 120 cm, l₂ = 120 – 20 = 100 cm
r = R(\(\frac{l_{1}-l_{2}}{l_{2}}\))
= 10 (\(\frac{120-100}{100}\))
= 2 Ω
The internal resistance of the cell is 2 Ω.

Question 17.
A potential drop per unit length along a wire is 5 × 10^-3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.
Answer:
Data: K = 5 × 10^-3 \(\frac{\mathrm{V}}{\mathrm{m}}\), L = 216 cm = 216 × 10^-2 m
E = KL
∴ E = 5 × 10^-3 × 216 × 10^-2
= 1080 × 10^-5
= 0.01080V
The emf of the cell is 0.01080 volt.
(Note: For K = 0.5 V/m, we get, E = 1.08V (reasonable value)]

Question 18.
The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1 µV/mm?
Answer:
Data: R = 8 Ω, L = 8 m, E = 2 V, K = 1 µV/mm
= 1 × \(\frac{10^{-6} \mathrm{~V}}{10^{-3} \mathrm{~m}}\) = 10^-3 \(\frac{\mathrm{V}}{\mathrm{m}}\)
K = \(\frac{V}{L}=\frac{E R}{\left(R+R_{\mathrm{B}}\right) L^{\prime}}\) where R_B is the resistance in the box.
∴ 10^-3 = \(\frac{2 \times 8}{\left(8+R_{B}\right) 8}\)
∴ 8 + R_B = \(\frac{2}{10^{-3}}\)
= 2 × 10³
∴ R_B = 2000 – 8
= 1992 Ω

Question 19.
Find the equivalent resistance between the terminals F and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm.

Answer:
Applying Kirchhoff’s voltage law to loop FGHF, we get,
– 10I₁ – 10(I₁ – I₂) + 10(I – I₁) + 10(I – I₁) = 0
∴ – 10I₁ – 10I₁ + 10I₂ + 10I – 10I₁ + 10I – 10I₁ = 0
∴ 201 – 40I₁ + 10I₂ = 0
∴ 2I – 4I₁ + I₂ = 0 …………….. (1)
Applying Kirchhoff’s voltage law to loop GABHG, we get,
– 10I₂ – 10I₂ + 10(I – I₂) + 10(I₁ – I₂) = 0
∴ – 20I₂ + 10I – 10I₂ + 10I₁ – 10I₂ = 0
∴ 10I + 10I₁ – 40I₂ = 0 .
∴ I + I₁ – 4I₂ = 0 ……………… (2)
Applying Kirchhoff’s voltage law to loop EFHBCDE, we get,
– 10(I – I₁) – 10(I – I₁) – 10(I – I₂) + E = 0
∴ -10I + 10I₁ – 10I + 10I₁ – 10I + 10I₂ + E = 0
∴E = 30I – 20I₁ – 10I₂ ………….. (3)
From Eq. (1), we get, I₂ = 4I₁ – 2I …………. (4)
From Eqs. (2) and (4), we get,
I + I₁ – 4(4I₁ – 2I) = 0
∴ I + I₁ – 16I₁ + 8I = 0 .
∴ 9I = 15I₁ ∴ I₁ = \(\frac{9}{15}\)I = \(\frac{3}{5}\)I …………. (5)
From Eqs. (4) and (5), we get,
I₂ = 4(\(\frac{3}{5}\)I) – 2I = \(\frac{12}{5}\)I – 2I
= \(\frac{12 I-10 I}{5}\) = \(\frac{2}{5}\) I
From Eqs. (3), (5) and (6), we get
E = 30I – 20(\(\frac{3}{5}\) I) – 10(\(\frac{2}{5}\) I)
= 30I – 12I – 4I = 30I – 16I
∴ E = 14I
If R is the equivalent resistance between E and C,
E = RI
∴ R = 14 Ω

Question 20.
A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Ω?
Answer:
Data: E = 2V, r = 20 Ω, R = 100 Ω
The voltmeter reading, V = IR
V = (\(\frac{2}{100+20}\))100 = \(\frac{200}{120}=\frac{10}{6}\) = 1.667 V.

12th Physics Digest Chapter 9 Current Electricity Intext Questions and Answers

Observe and Discuss (Textbook Page No. 220)

Question 1.
Post Office Box
A post office box (PO Box) is a practical form of Wheatstone bridge as shown in the figure.

It consists of three arms P, Q and R. The resistances in these three arms are adjustable. The two ratio arms P and Q contain resistances of 10 ohm, 100 ohm and 1000 ohm each. The third arm R contains resistances from 1 ohm to 5000 ohm. The unknown resistance X (usually, in the form of a wire) forms the fourth arm of the Wheatstone’s bridge. There are two tap keys K₁ and K₂ .
Answer:
The resistances in the arms P and Q are fixed to a desired ratio. The resistance in the arm R is adjusted so that the galvanometer shows no deflection. Now the bridge is balanced. The unknown resistance X = RQ/P, where P and Q are the fixed resistances in the ratio arms and R is the adjustable known resistance.

If L is the length of the wire used to prepare the resistor with resistance X and r is its radius, then the specific resistance (resistivity) of the material of the wire is given by
ρ = \(\frac{X \pi r^{2}}{L}\)

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e^-1 = 0.3678.
Solution:
∵ m = 1
∵ X follows Poisson Distribution

= e^-m × 1 + e^-m × 1
= e^-1 + e^-1
= 2 × e^-1
= 2 × 0.3678
= 0.7356

Question 2.
If X ~ P(\(\frac{1}{2}\)), then find P(X = 3) given e^-0.5 = 0.6065.
Solution:

Question 3.
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e^-3 = 0.0497
Solution:
∵ X follows Poisson Distribution

Question 4.
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e^-4 = 0.0183.
Solution:
∵ m = 1
∵ X ~ P(m = 4)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
X = No. of complaints recieved
(i) P(Only two complaints on a given day)

(ii) P(Atmost two complaints on a given day)
P(X ≤ 2) = p(0) + p(1) + p(2)
= \(\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}\) + 0.1464
= e^-4 + e^-4 × 4 + 0.1464
= e^-4 [1 + 4] + 0.1464
= 0.0183 × 5 + 0.1464
= 0.0915 + 0.1464
= 0.2379

Question 5.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given e^-1.5 = 0.2231.
Solution:
Let X = No. of demands for a car on any day
∴ No. of cars hired
n = 2
m = 1.5
∵ X ~ P(m = 1.5)

Question 6.
Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e^-1 = 0.3678.
Solution:
∵ X = No. of defects on a plywood sheet
∵ m = -1
∵ X ~ P(m = -1)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(No defect)
P(X = 0) = \(\frac{e^{-1} \times 1^{0}}{0 !}\)
= e^-1
= 0.3678

(ii) P(At least one defect)
P(X ≥ 1) = 1 – P(X < 1)
= 1 – p(0)
= 1 – 0.3678
= 0.6322

Question 7.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive. Given e^-5 = 0.0067.
Solution:
X = No. of rats
∵ m = 5
∴ X ~ P(m = 5)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(Exactly five rats)

(ii) P(More than five rats)
P(X > 5) = 1 – P(X ≤ 5)

(iii) P(between 5 and 7 rats, inclusive)
P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)

= 0.0067 × 3125 × 0.02
= 0.0067 × 62.5
= 0.42

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 1 Reproduction in Lower and Higher Plants Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

1. Multiple Choice Questions

Question 1.
Insect pollinated flowers usually possess ………………
(a) sticky pollen with a rough surface
(b) large quantities of pollens
(c) dry pollen with a smooth surface
(d) light-colored pollens
Answer:
(a) sticky pollen with a rough surface

Question 2.
In ovule, meiosis occurs in ………………
(a) Integument
(b) Nucellus
(c) Megaspore
(d) Megaspore mother cell
Answer:
(d) Megaspore mother cell

Question 3.
The ploidy level is NOT the same in ………………
(a) Integuments and nucellus
(b) Root tip and shoot tip
(c) Secondary nucleus and endosperm
(d) Antipodals and synergids
Answer:
(c) Secondary nucleus and endosperm

Question 4.
Which of the following types require pollination but result is genetically similar to autogamy?
(a) Geitonogamy
(b) Xenogamy
(c) Apogamy
(d) Cleistogamy
Answer:
(a) Geitonogamy

Question 5.
If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?
(a) Endosperm
(b) Leaf cells
(c) Cotyledons
(d) Synergids
Answer:
(d) Synergids

Question 6.
In angiosperms, endosperm is formed by/ due to ………………
(a) free nuclear divisions of megaspore
(b) polar nuclei
(c) polar nuclei and male gamete
(d) synergids and male gametes
Answer:
(c) polar nuclei and male gamete

Question 7.
Point out the odd one.
(a) Nucellus
(b) Embryo sac
(c) Micropyle
(d) Pollen grain
Answer:
(d) Pollen grain

2. Very Short Answer Questions

Question 1.
The part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigmatic surface.

Question 2.
How many haploid cells are present in a mature embryo sac?
Answer:
6 cells, 2 synergids, 1 egg cell, 3 antipodals.

Question 3.
Even though each pollen grain has 2 male gametes why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?
Answer:
Angiosperms have phenomenon of double fertilization in which both the male gametes are utilized, one for fusion with egg cell to form zygote and other for fusion with secondary nucleus to form endosperm.

Question 4.
Megasporogenesis
Answer:
It is the process of formation of haploid megaspores from diploid megaspore mother cell.

Question 5.
What is hydrophily?
Answer:
Transfer of pollen grains in pollination process through agency of water is known as hydrophily.

Question 6.
The layer which supplies nourishment to the developing pollen grains.
Answer:
Tapetum

Question 7.
Parthenocarpy
Answer:
The condition in which fruit is developed without the process of fertilization is called parthenocarpy.

Question 8.
Are pollination and fertilization necessary in apomixis?
Answer:
Apomixis is formation of embryos without formation of gametes hence there is no need of pollination and fertilization.

Question 9.
The part of pistil which develops into fruit and seed.
Answer:
Ovary develops into fruit and ovules into seed.

Question 10.
What is the function of filiform apparatus ?
Answer:
Filiform apparatus guides the pollen tube towards egg cell.

3. Short Answer Questions

Question 1.
How polyembryony can be commercially exploited?
Answer:

1. Polyembryony is the development of more than one embryo inside the seed.
2. When such polyembryonic seed germinate we get multiple seedlings from it.
3. This condition increases the chances of survival of new plants.
4. Nucellar embryos are genetically identical to parent plants hence we get uniform plants.
5. In horticulture we can utilize these as rootstock for grafting, hence they have significant role in fruit breeding programmes e.g. Citrus, Mango.

Question 2.
Pollination and seeds formation are very crucial for the fruit formation, Justify.
Answer:

1. After fertilization, ovary is transformed into fruit, where ovary wall becomes fruit wall, i.e pericarp.
2. Mature ovules are transformed into seeds after fertilization.
3. Fertilization is a process where male gametes unites with female gamete to form zygote which develops into embryo.
4. In pollination process pollen grains carrying non-motile male gamete are transferred on stigma.
5. Seeds have embryo which germinate into new plant hence the goal of reproduction to create offspring for next generation is achieved. Hence these are the crucial events for fruit formation.

Question 3.
Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?
Answer:

1. Self incompatibility or self-sterility is a genetic mechanism that prevents germination of pollen on stigma of same flower. This favours cross pollination. E.g. Tobacco.
2. In pollen-pistil interaction, when pollen grain is deposited on stigma, pistil has the ability to recognize and allow germination of right type of pollen.
3. Special type of proteins on stigmatic surface determine compatibility or incompatibility.
4. A physiological mechanism operates to ensure successful germination of compatible pollen.
5. Compatible pollen absorbs water and nutrients from stigmatic surface that are absent in pollen and then pollen tube emerges which grow-s through style.

Question 4.
Describe three devices by which cross pollination is encouraged in Angiosperms by avoiding self-pollination?
Answer:

1. Unisexuality, dichogamy, prepotency, heteromorphy and herkogamy are the outbreeding devices.
2. Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.
3. Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.
4. Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. E.g. Calotropis where pollinia are situated below the stigma.

4. Long Answer Questions

Question 1.
Describe the process of double fertilization.
Answer:
Double fertilization:
(1) Out of the two male gametes produced by the male gametophyte in angiosperms, one unites with the female gamete and the other with the secondary nucleus. Since both the male gametes take part in fertilization and fertilization occurs twice, it is called double fertilization.

(2) During double fertilization, the pollen tube on reaching the ovule enters the embryo sac through micropyle and bursts in one of the synergids. Owing to this, the two male gametes contained in the pollen tube, are set free.

(3) Out of the two male gametes, one unites with the egg or female gamete and the other unites with the secondary nucleus of the embryo sac, forming a triploid or triple fusion nucleus, called the primary endosperm nucleus. The process involving the fusion of one of the male gametes with the egg nucleus, resulting in the formation of a diploid zygote is called syngamy.

(4) The reproductive process in which non-motile male nuclei are carried to the egg cell through a pollen tube is called siphonogamy.

(5) After fertilization, zygote develops into an embryo. Certain changes take place in the ovule leading to the development of a seed.

Question 2.
Explain the stages involved in the maturation of microspore into male gametophyte.
OR
Describe the development of male gametophyte before pollination in angiosperms.
OR
Sketch and label male gametophyte in angiosperm.
Answer:

1. Microspore or pollen grain is first cell of male gametophyte.
2. The protoplast of pollen grain divides mitotically to form two unequal cells – a small thin walled generative cell and a large naked vegetative or tube cell.
3. The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.
4. The vegetative, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.
5. In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.
6. The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

Question 3.
Explain the development of dicot embryo.
Answer:
Development of embryo (dicot) in angio- sperm:

The oospore undergoes a transverse division to form a large basal cell towards the micropyle and a small apical or terminal cell towards the chalaza of the embryo sac. This two celled structure is called proembryo. The basal cell or suspensor initial undergoes repeated transverse divisions to form a multicellular structure called suspensor. The suspensor pushes the embryo towards the endosperm to draw its nutrition.

1. The development of embryo from a zygote is called embryogenesis.
2. The fusion of male gamete and an egg cell during fertilization results in the formation of a diploid zygote. The zygote develops a wall around it and is converted into oospore.
3. The apical cell or embryonal initial of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
4. From octant, the lower four cells form hypocotyl and radicle while four cells of upper side form plumule with two cotyledons.
5. The lowermost cell of suspensor is hypophysis and by its further division forms part of radicle and root cap.
6. The cells from upper side of octant divide repeatedly to form heart shaped which elongated further to form two lateral cotyledons.
7. Enlargement of hypocotyl and cotyledon results into curved embryo which appears horse shoe shaped.

Question 4.
Draw a diagram of the L.S of anatropous ovule and list the components of embryo sac and mention their fate after fertilization.
Answer:
Components of Embryo sac.

1. Mature embryo sac is 7-celled and 8 nucleate.
2. Egg apparatus at micropylar end – with 2 synergids and egg cell.
3. Central cell with secondary nucleus formed by 2 polar nuclei
4. Antipodal cells at chalazal end – 3 cells.
5. Pollen tube enters the synergids, Synergids guide the growth of pollen tube towards egg.
6. Male gamete fuses with female gamete, i.e. syngamy to form zygote which develops into embryo.
7. One male gamete fuses with secondarynucleus to form primary endosperm nucleus (PEN) which forms endosperm, nutritive tissue for embryo.

5. Fill in the Blanks

Question 1.
The ……………… collects the pollen grains.
Answer:
biotic agents

Question 2.
The male whorl, called the ……………… produces ………………
Answer:
androecium, pollen grains

Question 3.
The pollen grains represent the ………………
Answer:
male

Question 4.
The ……………… contains the egg or ovum.
Answer:
embryo sac

Question 5.
…………….. takes place when one male gamete and the egg fuse together. The fertilized egg grows vs into seed from which the new plants can grow.
Answer:
Fertilization

Question 6.
The ……………… is the base of the flower to which other floral parts are attached.
Answer:
thalamus

Question 7.
……………… is the transfer of pollen grains from anther of the flower to the stigma of the same or a different flower.
Answer:
Pollination

Question 8.
Once the pollen reaches the stigma, pollen tube traverses down the ……………… to the ovary where fertilization occurs.
Answer:
style

Question 9.
The ……………… are coloured to attract the insects that carry the pollen. Some flowers also produce ……………… or ……………… that attracts insects.
Answer:
petals, fragrance, nectar

Question 10.
The whorl ……………… is green that protects the flower until it opens.
Answer:
Calyx.

6. Label the Parts of seed.

Answer:

7. Match the following

Column I (Structure Before seed formation)	Column II (Structure After seed formation)
A. Funiculus	i. Hilum
B. Scar of Ovule	ii. Tegmen
C. Zygote	iii. Testa
D. Inner Integument	iv. Stalk of Seed
	v. Embryo

(a) A-v, B-i, C-ii, D-iv
(b) A-iii, B-iv, C-i, D-v
(c) A-iv, B-i, C-v D-ii
(d) A-iv, B-v C-iii, D-ii
Answer:
(c) A-iv, B-i, C-v D-ii

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 10 Halogen Derivatives Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 16 Green Chemistry and Nanochemistry

1. Choose the most correct option.

Question i.
The development that meets the needs of the present without compromising the ability of future generations to meet their own need is known as
a. Continuous development
b. Sustainable development
c. True development
d. Irrational development
Answer:
b. Sustainable development

Question ii.
Which of the following is ϒ-isomer of BHC?
a. DDT
b. lindane
c. Chloroform
d. Chlorobenzene
Answer:
b. lindane

Question iii.
The prefix ‘nano’ comes from
a. French word meaning billion
b. Greek word meaning dwarf
c. Spanish word meaning particle
d. Latin word meaning invisible
Answer:
(b) Greek word meaning dwarf

Question iv.
Which of the following information is given by FTIR technique?
a. Absorption of functional groups
b. Particle size
c. Confirmation of formation of nanoparticles
d. Crystal structure
Answer:
(a) Absorption of functional groups

Question v.
The concept of green chemistry was coined by
a. Born Haber
b. Nario Taniguchi
c. Richard Feynman
d. Paul T. Anastas
Answer:
(d) Paul T. Anastas

2. Answer the following

Question i.
Write the formula to calculate % atom economy.
Answer:

Question ii.
Name the ϒ-isomer of BHC.
Answer:
Lindane

Question iii.
Ridhima wants to detect structure of surface of materials. Name the technique she has to use.
Answer:
Scanning electron microscopy (SEM)

Question iv.
Which nanomaterial is used for tyres of car to increase the life of tyres?
Answer:
Carbon black

Question v.
Name the scientist who discovered scanning tunneling microscope (STM) in 1980.
Answer:
Gerd Binning and Heinrich Rohrer. (Nobel prize 1986)

Question vi.
1 nm = …..m?
Answer:
1 nm = 10⁹ m

3. Answer the following

Question i.
Define
(i) Green chemistry
(ii) sustainable development.
Answer:
(i) Green chemistry : Green chemistry is the use of chemistry for pollution prevention and it designs the use of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.

(ii) Sustainable development : Sustainable development is the development that meets the needs of the present, without compromising the ability of future generations to meet their own needs.

Question ii.
Explain the role of green chemistry.
Answer:
When the waste and pollution that society generates exceeds the Earth’s natural capacity for dealing with it, the green chemistry approach plays an important role.

• To reduce or eliminate the use or generation of hazardous substances in the design, manufacture and use of chemical products by promoting innovative chemical technologies.
• Capital expenditure required for prevention of pollution is controlled by the use of green chemistry.
• Since green chemistry incorporates and promotes pollution prevention practices in the manufacturing process of chemicals it helps industrial ecology.
• Green chemistry helps to protect the presence of ozone in the stratosphere. Ozone layer is essential for the survival of life on the earth.
• Global warming (Greenhouse effect) is controlled by green chemistry. At present it is the beginning of the green revolution.
• It is an exciting time with the new challenges for chemist involved with the discovery, manufacturing and use of chemicals. Green chemistry helps us to save environment and save earth, which is important for our future.

Question iii.
Give the full form (long form) of the names for the following instruments.
a. XRD
b. TEM.
c. STM
d. FTIR
e. SEM
Answer:
a. XRD-X-ray diffraction
b. TEM-Tunneling Electron Microscope
c. STM – Scanning Tunneling Microscope
d. FTIR-Fourier Transform Infrared Spectroscope
e. SEM-Scanning Electron Microscope

Question iv.
Define the following terms :
a. Nanoscience
b. Nanotechnology
c. Nanomaterial
d. Nanochemistry
Answer:
a. Nanoscience : The study of phenomena and manipulation of materials at atomic, molecular and macromolecular scales where properties differ significantly from those at a larger scale is called nanoscience.

b. Nanotechnology : The design, characterization, production and application of structures, device and system by controlling shape and size at nanometer scale is called nanotechnology.

c. Nanomaterial : A material having structural components with at least one dimension in the nanometer scale that is 1 -100 nm is called the nanomaterial. Nanomaterials are larger than single atoms but smaller than bacteria and cells.

d. Nanochemistry : It is the combination of chemistry and nanoscience. It deals with designing and synthesis of materials of nanoscale with different size and shape, structure and composition and their organization into functional architectures.

Question v.
How nanotechnology plays an important role in water purification techniques?
Answer:

1. Water purification is an important issue as 1.1 billion people do not have access to improved water supply. Water contains water bom pathogens like viruses, bacteria.
2. Silver nanoparticles are highly effective bacterial disinfectant to remove E. Coli from water. Hence, filter materials coated with silver nanoparticles is used to clean water.
3. Silver nanoparticles (AgNps) is a cost effective alternative technology (for e.g. water purifier).

Question vi.
Which nanomaterial is used in sunscreen lotion? Write its use.
Answer:
Zinc oxide (ZnO) and Titanium dioxide (TiO₂) nanoparticles are used sunscreen lotions. The chemicals protect the skin against harmful u.v (ultraviolet) rays by absorbing or reflecting the light and prevent the skin from damage.

Question vii.
How will you illustrate the use of safer solvent and auxiliaries?
Answer:

• Use of safer solvents and auxiliaries – is a principle of green chemistry it states that safer solvent like water, supercritical CO₂ should be used in place of volatile halogenated organic solvents, like CH₂CI₂, CHCI₃, CCI₄ for chemical synthesis and other purposes.
• Solvents dissolve solutes and form solutions, they facilitate many reactions. Water is a safer benign solvent while solvents like dichloromethane (CH₂CI₂), chloroform (CHCI₃) etc are hazardous.
• Use of toxic solvents affect millions of workers every year and have implications for consumers and the environment. A large amount of waste is created by their use and they also have huge environmental and health impacts.
• Finding safer solvents or designing processes which are solvent free is the best way to improve the process and the product.

Question viii.
Define catalyst. Give two examples.
Answer:
A substance which speeds up the rate of a reaction without itself being changed chemically in the reaction is called a catalyst. It helps to increase selectivity, minimise waste and reduce reaction time and energy demands. For example : Hydrogenation of oil the catalyst used are platinum or palladium, Raney nickel.

4. Answer the following

Question i.
Explain any three principles of green chemistry.
Answer:

1. Environment protection is the prime concern which has lead to the need for designing chemicals that degrade and can be discarded easily. These chemicals and their degradation products should be non-toxic, non-bioaccumulative or should not be environmentally persistent.
2. This principle aims at waste product being automatically degradable to clean the environment. Thus the preference for biodegradable polymers and pesticides.
3. To make the separation and segregation easier for the consumer an international plastic recycle mark is printed on larger items.
4. There is a dire need to develop improvised analytical methods to allow for real time, in process monitoring and control prior to the formation of hazardous substances.
5. It is very much important for the chemical industries and nuclear reactors to develop or modify analytical
   methodologies so that continuous monitoring of the manufacturing and processing unit is possible.
6. It is needed to develop chemical processes that are safer and minimize the risk of accidents. It is important to select chemical substances used in a chemical reaction in such a way that they can minimize the occurrence of chemical accidents, explosions, fire and emissions.
7. For example : Chemical process that works with the gaseous substances can lead to relatively higher possibilities of accidents including explosion as compared to the system working with nonvolatile liquid and solid substances.

Question ii.
Explain atom economy with suitable example.
Answer:
(1) Atom economy is a measure of the amount of atoms from the starting material that are present in the final product at the end of a chemical process. Good atom economy means most of the atoms of the reactants are incorporated in the desired products. Only small amount of waste is produced, hence lesser problem of waste disposal.

(2) The atom economy of a process can be calculated using the following formula.

The atom economy of the above reacijon is less than 50% and waste produced is higher.

Question iii.
How will you illustrate the principle, minimization of steps?
Answer:
(1) The technique of protecting or blocking group is commonly used in organic synthesis. Finally on completion of reaction deprotection of the group is required. This leads to unnecessary increase in the number of steps and decreased atom economy.

(2) The green chemistry principle aims to develop processes to avoid necessary steps i.e. (minimization of steps). When biocatalyst is used very often there is no need for protection of selective group. For example, conversion of m-hydroxyl benzaldehyde to m-hydroxybenzoic acid.

Question iv.
What do you mean by sol and gel? Describe the sol-gel method of preparation for nanoparticles.
Answer:
(1) Sol : Sols are dispersions of colloidal particles in a liquid. Colloids are solid particles with diameter of 1-100 nm.

(2) Gel : A gel is interconnected rigid network with pores of submicrometer dimensions and polymeric chains whose average length is greater than a micrometer.

(3) Sol-gel Process : A sol-gel process is an inorganic polymerisation reaction. It is generally carried out at room temperature, it includes four steps : Hydrolysis, polycondensation, drying and thermal decomposition. This method is widely used to prepare oxide materials.

The reactions involved in the sol-gel process are as follows :
MOR + H₂O → MOH + ROH (hydrolysis)
metal alkoxide
MOH + ROM → M-O-M + ROH (condensation)

• Formation of different stable solution of the alkoxide or solvated metal precursor.
• Gelation involves the formation of an oxide or alcohol-bridged network (gel) by a polycondensation reaction.
• Aging of the gel means during that period gel transforms into a solid mass.
• Drying of the gel involves removal of water and other volatile liquids from the gel network.
• Dehydration is achieved when the material is heated at temperatures up to 800°C.

Question v.
Which flower is an example of self-cleaning?
Answer:

• Lotus is an example of self cleansing.
• Nanostructures on the lotus plant leaves are super hydrophobic, they repel water which carries dirt as it rolls off.
  Thus though lotus plant (Nelumbonucifera) grows in muddy water, its leaves always appear clean.

Activity :
Collect information about the application of nanochemistry in cosmetics and pharmaceuticals

12th Chemistry Digest Chapter 16 Green Chemistry and Nanochemistry Intext Questions and Answers

Do you know? (Textbook page 343)

Question 1.
Does plastic packaging impact the food they wrap ?
Answer:
Phthalates leach into food through packaging so you should avoid microwaving food or drinks in plastic and not use plastic cling wrap and store your food in glass container whenever possible. Try to avoid prepackaging, processed food so that you will reduce exposure to the harmful effects of plastic.

Used Catalyst (Textbook page 342)

Question 18.
Complete the chart:

Reaction	Name of Catalyst used
1. Hydrogenation of oil (Hardening)	…………………………………
2. Haber’s process of manufacture of ammonia	…………………………………
3. Manufacture of HDPE polymer	…………………………………
4. Manufacture of H2S04 by contact process	…………………………………
5. Fischer-Tropsch process (synthesis of gasoline)	…………………………………

Answer:

Reaction	Name of Catalyst used
1. Hydrogenation of oil (Hardening)	Nickel (Ni)
2. Haber’s process of manufacture of ammonia	Iron
3. Manufacture of HDPE polymer	Zeigler-Natta catalyst
4. Manufacture of H2S04 by contact process	Vanadium oxide (V205)
5. Fischer-Tropsch process (synthesis of gasoline)	Cobalt-based or Iron based

 

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 15 Introduction to Polymer Chemistry

1. Choose the correct option from the given alternatives.

Question i.
Nylon fibers are …………………………………..
A. Semisynthetic fibres
B. Polyamide fibres
C. Polyester fibres
D. Cellulose fibres
Answer:
B. polyamide fibres

Question ii.
Which of the following is naturally occurring polymer?
A. Telfon
B. Polyethylene
C. PVC
D. Protein
Answer:
D. Protein

Question iii.
Silk is a kind of …………………………………. fibre
A. Semisynthetic
B. Synthetic
C. Animal
D. Vegetable
Answer:
C. an animal

Question iv.
Dacron is another name of …………………………………. .
A. Nylon 6
B. Orlon
C. Novolac
D. Terylene
Answer:
D. Terylene

Question v.
Which of the following is made up of polyamides?
A. Dacron
B. Rayon
C. Nylon
D. Jute
Answer:
C. Nylon

Question vi.
The number of carbon atoms present in the ring of ε – caprolactam is
A. Five
B. Two
C. Seven
D. Six
Answer:
D. Six

Question vii.
Terylene is …………………………………. .
A. Polyamide fibre
B. Polyester fibre
C. Vegetable fibre
D. Protein fibre
Answer:
B. Polyester fibre

Question viii.
PET is formed by …………………………………. .
A. Addition
B. Condensation
C. Alkylation
D. Hydration
Answer:
D. Hydration

Question ix.
Chemically pure cotton is …………………………………. .
A. Acetate rayon
B. Viscose rayon
C. Cellulose nitrate
D. Cellulose
Answer:
D. Cellulose

Question x.
Teflon is chemically inert, due to presence of …………………………………. .
A. C-H bond
B. C-F bond
C. H- bond
D. C=C bond
Answer:
A. C-H bond

2. Answer the following in one sentence each.

Question i.
Identify ‘A’ and ‘B’ in the following reaction …………………………………. .

Answer:

Question ii.
Complete the following statements
a. Caprolactam is used to prepare …………………………………. .
b. Novolak is a copolymer of …………………………………. and …………………………………. .
c. Terylene is ………………………………….. polymer of terephthalic acid and ethylene glycol.
d. Benzoyl peroxide used in addtion polymerisation acts as …………………………………. .
e. Polyethene consists of polymerised …………………………………. .
Answer:
a. Nylon-6
b. Phenol, formaldehyde
c. polyester
d. initiator (catalyst)
e. linear or branched-chain

Question iii.
Draw the flow chart diagram to show the classification of polymers based on type of polymerisation.
Answer:

Question iv.
Write examples of Addition polymers and condensation polymers.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene, Nylon-66

Question v.
Name some chain-growth polymers.
Answer:
Chain growth polymers : Polythene, polyacrylonitrile and polyvinyl chloride.

Question vi.
Define the terms :
1) Monomer
2) Vulcanisation
3) Synthetic fibres
Answer:
1. Monomer is a small and simple molecule and has a capacity to form two chemical bonds with other monomers. Examples : Ethene, Propylene.
2. The process by which a network of cross-links is introduced into an elastomer is called vulcanisation or it can also be defined as the process of heating natural rubber with sulphur to increase the tensile strength, toughness and elasticity of natural rubber is known as vulcanization of rubber.
3. The man-made fibres prepared by polymerization of one monomer or copolymerization of two or more monomers are called synthetic fibres.

Question vii.
What type of intermolecular force leads to high-density polymer?
Answer:
High density polymers have low degree of branching along the hydrocarbon chain. The molecules are closely packed together during crystallization. This closer packing means that the van der Waals attraction between the chains are greater and so the plastic (high density polymer) is stronger and has a melting point.

Question viii.
Give one example each of copolymer and homopolymer.
Answer:
Homopolymer : PVC, Nylon-6
Copolymer : Terylene, Buna-S

Question ix.
Identify Thermoplastic and Thermosetting Plastics from the following …………………………………. .
1. PET
2. Urea-formaldehyde resin
3. Polythene
4. Phenol formaldehyde
Answer:
Thermoplastic plastics : PET, Polythene
Thermosetting plastics : Urea formaldehyde resin, Phenol formaldehyde

3. Answer the following.

Question i.
Write the names of classes of polymers formed according to intermolecular forces and describe briefly their structural characteristics.
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question ii.
Write reactions of formation of :
a. Nylon 6
b. Terylene
Answer:
Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

• Terylene has relatively high melting point (265 °C)
• It is resistant to chemicals and water.

Uses :

• It is used for making wrinkle free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
• PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
• It is used for making many articles like bottles, jams, packaging containers.

Question iii.
Write the structure of natural rubber and neoprene rubber along with the name and structure of thier monomers.
Answer:

Question iv.
Name the polymer type in which the following linkage is present.
Answer:
The linkage is present in terylene or dacron polymer.

Question v.
Write the structural formula of the following synthetic rubbers :
a. SBR rubber
b. Buna-N rubber
c. Neoprene rubber
Answer:

Question vi.
Match the following pairs :
Name of polymer – Monomer
1. Teflon – a. CH₂ = CH₂
2. PVC – b. CF₂ = CF₂
3. Polyester – c. CH₂ = CHCl
4. Polythene – d. C₆H₅OH and HCHO
5. Bakelite – e. Dicarboxylic acid and polyhydoxyglycol
Answer:

1. Teflon – CF₂ = CF₂
2. PVC – CH₂ = CHCI
3. Polyester-Dicarboxylic acid and polyhydoxyglycol
4. Polythene – CH₂ = CH₂
5. Bakelite – C₆H₅OH and HCHO

Question vii.
Draw the structures of polymers formed from the following monomers
1. Adipic acid + Hexamethylenediamine
2. e – Aminocaproic acid + Glycine
Answer:

Question viii.
Name and draw the structure of the repeating unit in natural rubber.
Answer:
Repeating unit of natural rubber (Basic unit : isoprene)

Question ix.
Classify the following polymers as natural and synthetic polymers
a. Cellulose
b. Polystyrene
c. Terylene
d. Starch
e. Protein
f. Silicones
g. Orlon (Polyacrylonitrile)
h. Phenol-formaldehyde resins
Answer:

Natural Polymers	1. Cellulose 4. Starch 5. Protein
Synthetic Polymers	2. Polystyrene 3. Terylene 6. Silicones 7. Orion (Polyacrylonitrile) 8. phenol-formaldehyde resin

Question x.
What are synthetic resins? Name some natural and synthetic resins.
Answer:
Synthetic resins are artificially synthesised high molecular weight polymers. They are the basic raw material of plastic. The main properties of plastic depend on the synthetic resin it is made from.

Examples of natural resins : Rosin, Damar, Copal, Sandarac, Amber, Manila
Examples of synthetic resins : Polyester resin, Phenolic resin, Alkyl resin, Polycarbonate resin, Polyamide resin, Polyurethane resin, silicone resin, Epoxy resin, Acrylic resin.

Question xi.
Distinguish between thermosetting and thermoplastic resins. Write example of both the classes.
Answer:

Thermosetting resin	Thermoplastic resin
(1) They harden when heated. Once hardened it no longer melts.	(1) They soften when heated and harden again when cooled.
(2) They cannot be re-shaped.	(2) They can be reshaped
(3) They are strong, hard.	(3) They are weak, soft.
(4) Thermosetting resin show cross-linking. Examples : Melamine resin Epoxy resins, Bake-lite.	(4) Thermoplastic molecules do not cross link, hence are flexible. Examples : Polythene, polypropylene, nylon, polyester.

Question xii.
Write name and formula of raw material from which bakelite is made.
Answer:
The raw material or monomers used to prepare bakelite are o-hydroxymethyl phenol and formaldehyde (HCHO)

4. Attempt the following :

Question i.
Identify condensation polymers and addition polymers from the following.

Answer:

Question ii.
Write the chemical reactions involved in the manufacture of Nylon 6, 6
Answer:
Nylon-6, 6 is a linear polyamide polymer formed by the condensation polymerisation reaction. The monomers used in the preparation of Nylon-6, 6 are :
(1) Adipic acid : HOOC-(CH₂)₄-COOH
(2) Hexamethylene diamine : H₂N-(CH₂)₆-NH₂

When equimolar aqueous solutions of adipic acid and hexamethylene diamine are mixed and heated, there is neutralization to form a nylon salt. During polymerisation at 553 k nylon salt loses a water molecule to form nylon 6, 6 polymer. Both monomers (hexamethylene diamine and adipic acid) contain six carbon atoms each, hence the polymer is termed as Nylon-6,6.

Properties and uses : Nylon 6,6 is high molecular mass (12000 – 50000 u) linear condensation polymer. It possesses high tensile strength. It does not soak in water. It is used for making sheets, bristles for brushes, surgical sutures, textile fabrics, etc.

Question iii.
Explain the vulcanisation of rubber. Which vulcanizing agents are used for the following synthetic rubber.
a. Neoprene
b. Buna-N
Answer:
The process by which a network of cross links is introduced into an elastomer is called vulcanization.

Vulcanization enhances the properties of natural rubber like tensile strength, stiffness, elasticity, toughness etc. Sulphur forms cross links between polyisoprene chains which results in improved properties of rubber.

• For neoprene vulcanizing agent is MgO.
• For Buna-N vulcanizing agent is sulphur.

Question iv.
Write reactions involved in the formation of …………………………………. .
1) Teflon
2) Bakelite
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

2. Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

• Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
• C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
  Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question v.
What is meant by LDP and HDP? Mention the basic difference between the same with suitable examples.
Answer:

• LDP is low density polyethylene and HDP is high density polyethylene.
• LDP is a branched polymer with low density due to chains are loosely held and HDP is a linear polymer with density due to close packing.
• HDP is much stiffer than LDP and has high tensile strength and hardness.

LDP is mainly used in preparation of pipes for agriculture, irrigation and domestic water line connections. HDP is used in manufacture of toys and other household articles like bucket, bottles, etc.

Question vi.
Write preparation, properties and uses of Teflon.
Answer:
Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

• Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
• C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
  Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question vii.
Classify the following polymers as straight-chain, branched-chain and cross-linked polymers.

Answer:

5. Answer the following.

Question i.
How is polythene manufactured? Give their properties and uses.
Answer:
LDP means low density polyethylene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

• LDP films are extremely flexible, but tough chemically inert and moisture resistant.
• It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

• LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
• It is also used in submarine cable insulation.
• It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

• HDP is crystalline, melting point in the range of 144 – 150 °C.
• It is much stiffer than LDP and has high tensile strength and hardness.
• It is more resistant to chemicals than LDP.

Uses of HDP :

• HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
• It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question ii.
Is synthetic rubber better than natural rubber? If so, in what respect?
Answer:
Yes. Synthetic rubber is more resistant to abrasion than natural rubber and is also superior in resistance to heat and the effects of aging (lasts longer). Many types of synthetic rubber are flame-resistant, so they can be used in insulation for electrical devices.

It also remains flexible at low temperatures and is resistant to grease and oil. It is resistant to heat, light and certain chemicals.

Question iii.
Write main specialities of Buna-S, Neoprene rubber?
Answer:
Buna-S is an elastomer and it is copolymer of styrene with butadiene. Its trade name is SBR. Buna-S is superior to natural rubber, because of its mechanical strength and abrasion resistance. It is used in tyre industry. It is vulcanized with sulphur. Neoprene is a synthetic rubber and it is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene). Vulcanization of neoprene takes place in presence of MgO. It is resistant to petroleum, vegetable oils. Neoprene is used in making hose pipes for transport of gasoline and making gaskets.

Question iv.
Write the structure of isoprene and the polymer obtained from it.
Answer:

Question v.
Explain in detail the free radical mechanism involved during the preparation of the addition polymer.
Answer:
Polymerisation of ethylene is carried out at high temperature and at high pressure in presence of small amount of acetyl peroxide as initiator.

(1) Formation of free radicals : The first step involves clevage of acetyl peroxide to form two carboxy radicals. These carboxy radicals immediately undergo decarboxylation to give methyl initiator free radicals.

(2) Chain initiating step : The methyl radical thus formed adds to ethylene to form a new larger free radical.

(3) Chain propagation step : The larger free radical formed in the chain initiating step reacts with another molecule of ethene to form another big size free radicals and chain grows. This is called chain propagation step.

The chain reaction continues till thousands of ethylene molecules are added.

(4) Chain terminating step : The continuous chain reaction can be terminated by the combination of free radicals to form polyethene.

Activity :
i. Collect the information of the process like extrusion and moulding in Textile Industries.
ii. Make a list of polymers used to make the following articles
a. Photographic film
b. Frames of spectacles
c. Fountain pens
d. Moulded plastic chains
e. Terywool or Terycot fabric
iii. Prepare a report on factors responsible for degradation of polymers giving suitable example.
iv. Search and make a chart/note on silicones with reference to monomers, structure, properties and uses.
v. Collect the information and data about Rubber industry, plastic industry and synthetic fibre (rayon) industries running in India.

12th Chemistry Digest Chapter 15 Introduction to Polymer Chemistry Intext Questions and Answers

Use your brain power! (Textbook Page No 323)

Question 1.
Differentiate between natural and synthetic polymers.
Answer:

Natural polymers	Synthetic polymers
(1) The polymers are obtained either from plants or animals.	(1) The man made fibres prepared by polymerization of monomer or copolymerization of two or more monomers.
(2) They are further divided into two types : (i) plant polymers (ii) Animal polymers.Examples: Cotton, linen, latex	(2) They are further divided into three subtypes : (i) fibres (ii) synthetic rubbers (iii) plastics.Examples : Nylon, terylene Buna-S

Use your brain power! (Textbook Page No 325)

Question 1.
What is the type of polymerization in the following examples?

Answer:
(i) Addition polymerization
(ii) Condensation polymerization

Problem 15.1 : (Textbook Page No 326)

Question 1.
Refer to the following table listing for different polymers formed from respective monomers. Identify from the list whether it is copolymer or homopolymer.

Monomer	Polymers
Ethylene	Polyethene
Vinyl chloride	Polyvinyl chloride
Isobutylene	Polyisobutylene
Acrylonitrile	Polyacrylonitrile
Caprolactam	Nylon 6
Hexamethylene diammonium adipate	Nylon 6, 6
Butadiene + styrene	Buna-S

Solution :
In each of first five cases, there is only one monomer which gives corresponding homopolymer. In the sixth case hexamethylene diamine reacts with adipic acid to form the salt hexamethylene diammonium adipate which undergoes condensation to form Nylon 6, 6. Hence nylon 6, 6 is homopolymer. The polymer Buna-S is formed by polymerization of the monomers butadiene and styrene in presence of each other. The repeating units corresponding to the monomers butadiene and styrene are randomly arranged in the polymer. Hence Buna-S is copolymer.

Use your brain power! (Textbook Page No 328)

Question 1.
(1) From the cis-polyisoprene structure of natural rubber explain the low strength of van der Waals forces in it.
(2) Explain how the vulcanization of natural rubber improves its elasticity. (Hint : consider the intermolecular links.)
Answer:
(1) (i) Natural rubber is cis-polyisoprene. It is obtained by polymerization of isoprene units at 1, 4 positions. In rubber molecule, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chain to come closer. Therefore, only weak vander Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

(ii) Cis-1, 4 polyisoprene (Natural rubber), due to this cis configuration about the double bonds, has the adjacent chain that do not fit together well (there is no close packing of adjacent chains). The only force that interact is the weak or low strength of van der Waals’ forces.

(iii) Cis-polyisoprene has a coiled structure in which the various polymer chains are held together by weak van der Waals’ forces.

(2) (i) Vulcanization of rubber is a process of improvement of the rubber elasticity and strength by heating it in the presence of sulphur, which results in three dimensional cross-linking of the chain rubber molecules (polyisoprene) bonded to each other by sulphur atoms.

(ii) Vulcanisation makes rubber more elastic and more stiffer. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus rubber get stiffened.

(iii) The improved properties of vulcanised rubber are (i) high elasticity (ii) low water absorption tendency,

(iii) resistance to oxidation.

Use your brain power! (Textbook Page No 334)

Question 1.
Write structural formulae of styrene and polybutadiene.
Answer:

(1) Classify the following polymers as addition or condensation.
(i) PVC (ii) Polyamides
(iii) Polystyrene
(iv) Polycarbonates
(v) Novolac
Answer:
Addition polymers: PVC, Polystyrene
(ondensatlon polymers: Polyamides. Polycarbonates, Novolac

Question 2.
Completed the following table :
Answer:

Use your brain power! (Textbook Page No 335)

(1) Represent the copolymerization reaction between glycine and e aminocaproic acid to form the copolymer nylon 2-nylon 6.
(2) What is the origin of the numbers 2 and 6 in the name of this polymer?
Answer:
(1) It is a copolymer and has polyamide linkages. The monomers glycine and e-amino caproic acid undergo condensation polymerisation to form nylon-2-nylon-6.

Nylon-2-nylon-6 is used in orthopaedic devices and implants.

(2) Monomer glycine contains two carbon atoms and e amino caproic acid contains six carbon atoms, hence the polymer is termed as nylon-2-nylon-6.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

1. Select the most correct choice.

Question i.
CH₂OH-CO-(CHOH)₄-CH₂OH is an example of
a. Aldohexose
b. Aldoheptose
c. Ketotetrose
d. Ketoheptose
Answer:
(d) Ketoheptose

Question ii.
The open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group
Answer:
(b) Anomeric hydroxyl group

Question iii.
Which of the following does not apply to CH₂NH₂ – COOH
a. Neutral amino acid
b. L – amino acid
c. Exists as zwitterion
d. Natural amino acid
Answer:
(d) Natural amino acid

Question iv.
Tryptophan is called essential amino acid because
a. It contains an aromatic nucleus.
b. It is present in all the human proteins.
c. It cannot be synthesized by the human body.
d. It is an essential constituent of enzymes.
Answer:
(c) It cannot be synthesised by human body.

Question v.
A disulfide link gives rise to the following structure of protein.
a. Primary
b. Secondary
c. Tertiary
d. Quaternary
Answer:
(c) Tertiary

Question vi.
RNA has
a. A – U base pairing
b. P – S – P – S backbone
c. double helix
d. G – C base pairing
Answer:
(a) A – U base pairing

2. Give scientific reasons :

Question i.
The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
Answer:
(1) In disaccharide sucrose, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a nonreducing sugar. As there is no free aldehyde group, it does not reduce Tollen’s reagent to metallic silver. Hence, sucrose gives negative Tollen’s test.

(2) While the disaccharide maltose is a reducing sugar because a free aldehyde group can be produced at C₁ of second sugar molecule. It is a reducing sugar. It reduces Tollen’s reagent to shining silver mirror. Hence, Maltose gives positive Tollen’s test.

Question ii.
On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.
Answer:
On complete hydrolysis DNA yields 2-deoxy-D-ribose, adenine, thymine, guanine, cystosine and phosphoric acid. Since adenine always forms two hydrogen bonds with thymine, the hydrolysis of DNA gives equimolar quantities of adenine and thymine.

Question iii.
α – Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.
Answer:

α-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, proton transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called zwitter ion.

Question iv.
Hydrolysis of sucrose is called inversion.
Answer:

Sucrose is dextro rotatory. On hydrolysis it gives equimolar mixture of D – ( + ) glucose and D – ( -) fructose. Since the laevorotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.7°), the hydrolysis product has net laevorotation. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro ( + ) to laevo (-) and the product is called as invert sugar and so the hydrolysis of sucrose is called inversion.

Question v.
On boiling, egg albumin becomes opaque white.
Answer:
Upon boiling the egg, denaturation αcurs. During denaturation, secondary and tertiary structures are destroyed, but primary structure remains intact. Egg contains soluble globular proteins, which forms insoluble fibrous proteins (opque) on boiling egg.

3. Answer the following

Question i.
Some of the following statements apply to DNA only, some to RNA only and some to both. Lable them accordingly.
a. The polynucleotide is double stranded. ( …………… )
b. The polynucleotide contains uracil. ( …………… )
c. The polynucleotide contains D-ribose ( …………… ).
d. The polynucleotide contains Guanine ( …………… ).
Answer:
(1) The polynucleotide is double stranded. (DNA)
(2) The polynucleotide contains uracil. (RNA)
(3) The polynucleotide contain D-ribose (RNA)
(4) Thc polynucleotide contains Guanine (DNA, RNA)

Question ii.
Write the sequence of the complementary strand for the following segments of a DNA molecule.
a. 5′ – CGTTTAAG – 3′
b. 5′ – CCGGTTAATACGGC – 3′
Answer:
(1) DNA molecule : 5′ – CGTTTAAG – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

(2) DNA molecule : 5′ – CCGGTTAATACGGC – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

Question iii.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Answer:
(1) Dipeptide formed from alanine and glycine

(2) Dipeptide formed from alanine and tyrosine

(3) Dipeptide formed from glycine and tyrosine.

Question iv.
Give two pieces of evidence for the presence of the formyl group in glucose.
Answer:
(1) Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of – CHO (formyl group) in glucose.

(2) Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid which shows carbonyl group in glucose is aldehyde (formyl group) group.

4. Draw a neat diagram for the following:

Question i.
Haworth formula of glucopyranose
Answer:

Question ii.
Zwitter ion
Answer:

Question iii.
Haworth formula of maltose
Answer:

Question iv.
Secondary structure of the protein

Answer:
The structure of proteins can be studied at four different levels i.e. primary, secondary, tertiary and quaternary levels. Each level is more complex than the previous one.
(1) Primary structure of proteins :
(a) Representation by structural formula

(b) Representation with amino acid symbols

Primary structure of proteins is the sequence of constituent a-amino acid residues linked by peptide bonds. Any change in the sequence of amino acid residue creates different protein molecule. Primary structure of proteins is represented by writing the three letter symbols of amino acid residues as per their sequence in the concerned protein. The symbols are separated by dashes. According to the convention, the N-terminal amino acid residue as written at the left end and the C-terminal amino acid residue at the right end.

(2) Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :

(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :

• The C = O and N – H bonds lie in the planes of the sheet.
• Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
• The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

(3) Tertiary structure of proteins :

The three-dimensional shape acquired by the entire polypeptide chain of a protein is called its tertiary structure. The structure is stabilized and has attractive interaction with the aqueous environment of the cell due to the folding of the chain in a particular manner. Tertiary structure gives rise to two major molecular shapes i.e. globular and fibrous proteins. The main forces which stabilize a particular tertiary structure include hydrogen bonding, dipole-dipole attraction (due to polar bonds in the side chains), electrostatic attraction (due to the ionic groups like -COO^–, \(\mathrm{NH}_{3}^{+}\) in the side chain) and also London dispersion forces. Finally, disulfide bonds formed by oxidation of nearby – SH groups (in cysteine residues) are the covalent bonds which stabilize the tertiary structure.

(4) Quaternary structure of proteins The two or more polypeptide chains with folded tertiary structures forms complex protein. The spatial arrangements of these polypeptide chains with respect to each other is known as quaternary structure. Each individual polypeptide chain is called a subunit of the overall protein. For example: Haemoglobin consists of four subunits called haeme held together by intermolecular forces in a compact three dimensional shape. Haemoglobin can do its function of oxygen transport only when all the four subunits are together.

Question v.
AMP
Answer:

Question vi.
dAMP
Answer:

Question vii.
One purine base from nucleic acid
Answer:

Question viii.
Enzyme catalysis
Answer:

Activity :

• Draw the structure of a segment of DNA comprising at least ten nucleotides on a big chart paper.
• Make a model of DNA double stranded structure as group activity.

12th Chemistry Digest Chapter 14 Biomolecules Intext Questions and Answers

Try ….. this (Textbook Page No 298)

Question 1.
Observe the following structural formulae carefully and answer the questions.

(1) How many OH groups are present in glucose, fructose and ribose respectively?
(2) Which other functional groups are present in these three compounds?
Answer:
(1) Glucose contains five hydroxyl (- OH) groups.
Fructose contains five hydroxyl ( – OH) groups.
Ribose contains four hydroxyl ( – OH) groups.

(2) Glucose contains aldehyde ( – CHO) as other functional group.
Fructose contains ketonic group as other functional group.
Ribose contains aldehyde ( – CHO) as other functional group.

Use your brain power! (Textbook Page No 299)

Question 1.
Give IUPAC names to the following monosaccharides.

Answer:
(1) Aldotriose
(2) Aldopentose
(3) Ketoheptose

Problem 14.1 : (Textbook Page No 300)

Question 1.
An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (- OH) groups must be present in the original compound?
Solution :
In acetylation reaction H atom of an (- OH) group is replaced by an acetyl group (- COH₃).

This results in an increase in molecular mass by [(12 + 16 + 12 + 3 x 1) – 1] that has, 42 u. In the given alcohol, increase in molecular mass = 174 u – 90 u = 84 u
∴ Number of – OH groups \(=\frac{84 \mathrm{u}}{42 \mathrm{u}}=2\)

Use your brain power! (Textbook Page No 301)

(1) Write structural formula of glucose showing all the bonds in the molecule.
(2) Number all the carbons in the molecules giving number 1 to the ( – CHO) carbon.
(3) Mark the chiral carbons in the molecule with asterisk (*).
(4) How many chiral carbons are present in glucose?
Answer:
Refer structural formula of glucose for (1) (2) and (3).

(4) There are 4 chiral carbon atoms present in glucose.

Use your brain power! (Textbook Page No 306)

Question 1.
(1) Is galactose an aldohexose or a ketohexose?
(2) Which carbon in galactose has different configuration compared to glucose?
(3) Draw Haworth formulae of α-D-galactose and β-D-galactose.
(4) Which disaccharides among sucrose, maltose and lactose is/are expected to give positive Fehling test?
(5) What are the expected products of hydrolysis of lactose?
Answer:

1. Galactose is an aldohexose.
2. Fourth carbon in galactose has different configuration compared to glucose.
3. 
4. Maltose and lactose are expected to give positive Fehling solution test.
5. The expected products of hydrolysis of lactose are D – ( +) glucose and D – ( +) galactose.

Can you think? (Textbook Page No 307)

Question 1.
When you chew plain bread, chapati or bhaakari for long time, it tastes sweet. What could be the reason?
Answer:
When chapati, bread or bhakari are chewed for long time the pulp mixes with saliva and carbohydrate component in them diseminates and gives the sweet taste.

Use your brain power! (Textbook Page No 309)

Question 1.
Tryptophan and histidine have the structures (I) and (II) respectively. Classify them into neutral? acidic/basic &amino acids and justify your answer. (Hint: Consider învolvement of lone pair in resonance).

Answer:

In tryptophan, nitrogen atom present in cyclic structure cannot donate pair of electrons as it is stabilized by resonance. The other amino group and carboxylic group present in the side chain neutralize each other. Tryptophan has equal number of amino and carboxylic groups. Hence, tryptophan is a neutral amino acid.

In histidine, amino groups are more in number than carboxyl groups therefore histidine ¡s basic in nature.

Can you think? (Textbook Page No 309)

Question 1.
Compare the molecular masses of the following compounds and explain the observed melting points.

Answer:
Above compounds have same molecular masses but they have different melting points, a-amino acids have higher melting points compared to the corresponding amines or carboxylic acids of comparable masses. This property is due to the presence of both carboxylic group (acidic) and amino group (basic) in the molecule. In aqueous solution, protons transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called – Zwitter ion.

Use your brain power! (Textbook Page No 310)

Question 1.
(1) Write the structural formula of dipeptide formed by combination of carboxyl group of alanine and amino group of glycine.
(2) Name the resulting dipeptide.
(3) Is this dipeptide same as glycyalanine or its structural isomer?
Answer:

(2) ala-glycine. OR ala-gly
(3) It is a structural isomer.

Question 54.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.

Problem 14.3 : (Textbook Page No 311)

Question 1.
Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine, tyrosine or tryptophan. Write the symbols of the amino acids and peptides smaller than pentapeptide formed by hydrolysis of the following hexapeptide with chymotrypsin. Gly-Tyr-Gly-Ala-Phe-Val
Solution :
In the given hexapeptide hydroylsis by chymotripsin can take place at two points, namely, Phe and Tyr. The carbonyl group of these residues is towards the right side, that is, toward the C-terminal. Therefore the hydrolysis products in required range will be :

Problem 14.4 : (Textbook Page No 311)

Question 1.
Write down the structures of amino acids constituting the following peptide.

Solution :
The given peptide has two amide bonds linking three amino acids. The structures of these amino acids are obtained by adding one H₂O molecule across the amide bond as follows :

Use your brain power! (Textbook Page No 313)

A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these giving rise to tertiary level structure of protein.

Answer:
Tertiary level structure from amino residues.

Can you tell? (Textbook Page No 313)

Question 1.
What is the physical change observed when (a) egg is boiled, (b) milk gets curdled on adding lemon juice?
Answer:
(a) When egg is boiled, coagulation of eggwhite (insoluble fibrous proteins) takes place. This is a common example of denaturation.
(b) When lemon juice is added to milk, it gets curdled due to the formation of lactic acid. This is another example of denaturation.

Can you tell? (Textbook Page No 315)

Question 1.
What is the single term that answers all the following questions?
(1) What decides whether you are blue eyed or brown eyed?
(2) Why does wheat grain germinate to produce wheat plant and not rice plant?
(3) Which acid molecules are present in nuclei of living cells?
Answer:
(1) Nucleic acid (DNA)
(2) Nucleic acid (DNA)
(3) Nucleic acid (DNA + RNA)

Use your brain power! (Textbook Page No 317)

Question 1.
Draw structural formulae of nucleosides formed from the following sugars and bases.
(1) D – ribose and guanine
(2) D – 2 – deoxyribose and thymine
Answer:
(1) D-ribose and guanine

(2) D – 2 – deoxyribose and thymine

Problem 14.5 (Textbook Page No 318)

Queston 1.
Draw a schematic representaion of trinucicotide segment ACT of a DNA molecule.
Solution :
In DNA molecule sugar is deoxyribose. The base ‘A’ in the given segment is at 5 end while the base T at the 3’ end. I-fence the schematic representation of the given segment of DNA is

Problem 14.6 : (Textbook Page No 320)

Question 1.
Write the sequence of the complementary strand of the following portion of a DNA molecule : 5 -ACGTAC-3
Solution :
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

Problem 14.2 : (Text Page No 303)

Question 1.
Assign D/L configuration to the following monosaccharides.

Solution :
D/L configuration is assigned to Fischer projection formula of monosaccharide on the basis of the lowest chiral carbon.

Threose has two chiral carbons C-2 and C-3. The given Fischer projection formula of threose has OH groups at the lowest C -3 chiral carbon on the right side.
∴ It is D-threose.

Ribose has three chiral carbons C – 2, C – 3 and C -4.
The given Fischer projection formula of ribose has – OH group at the lowest C -4 chiral carbon on the left side.
∴ It is L-ribose

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 13 Amines

1. Choose the most correct option.

Question i.
The hybridization of nitrogen in primary amine is ………………………..
a. sp
b. sp²
c. sp³
d. sp³d
Answer:
c. sp³

Question ii.
Isobutylamine is an example of ………………………..
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine

Question iii.
Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine

Question iv.
Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine

Question v.
Which type of amine does produce N₂ when treated with HNO₂?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine

Question vi.
Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine

Question vii.
Which one of the following compounds does not react with acetyl chloride?
a. CH₃-CH₂-NH₂
b. (CH₃-CH₂)₂NH
c. (CH₃-CH₂)₃N
d. C₆H₅-NH₂
Answer:
Answer:
c. (CH₃ – CH₂)₃N

Question viii.
Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine

Question ix.
Identify ‘B’ in the following reactions

Answer:
d. CH₃-CH₂-OH

Question x.
Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene

2. Answer in one sentence.

Question i.
Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:

Question ii.
How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.

Question iii.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH₃ – CH₂ – CONH₂) produces ethanamine by Hofmann bromamide degradation reaction.

Question iv.
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH₃.

Question v.
Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone f pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.

Question vi.
Predict the product of the following reaction. Nitrobenzene Sn/Conc. HCl?
Answer:
The product is aniline/

Question vii.
Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.

Question viii.
Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine

Question ix.
Write the balanced chemical equations for the action of dil H₂SO₄ on diethylamine.
Answer:

Question x.
Arrange the following amines in the increasing order of their _pK_b values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (_pK_A 3.34), aniline (_pK_A 9.13) 4-nitroaniline (_pK_A 12.99)

3. Answer the following

Question i.
Identify A and B in the following reactions.

Answer:

Question ii.
Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of K_b or _pK_b value. According to Lowry-Bron-sted theory the basic nature of amines is explained by the following equilibrium equation.

In this equilibrium amine accepts H⁺, hence an amine is a Lowry-Bronsted base.

According to Lewis theory, the species which donates a pair of electrons is called a base.

The nitrogen atom in amiqes has a lone pair of electrons, which can be donated to suitable acceptor like proton H⁺.

The aqueous solutions of amines are basic in nature due to release of free OH^– ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :

R – NH₂ + H₂O ⇌ RNH₃ + OH^–

Since OH^– is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH^–.

Here, K_b value is smaller and _pK_b value is larger.

Hence amines are weak bases.

Question iii.
What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 – 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

Diazotisation of aniline :

Question iv.
Write reaction to convert acetic acid into methylamine.
Answer:

Question v.
Write a short note on coupling reactions.
Answer:

Reactions involving retention of diazo group : (Coupling reactions) :

Question vi.
Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.

Question vii.
Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.

Secondary and tertiary amines do not give this test.

Question viii.
Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine

(2) Aniline into p-bromo aniline

Question ix.
Complete the following reactions :
a. C₆H₅N₂ Cl + C₂H₅OH →
b. C₆H₅NH₂ + Br₂(aq) → ?
Answer:
(a)

(b)

Question x.
Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C – X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylam- monium bromide (a quaternary ammonium salt).

The order of reactivity of alkyl halides with ammonia is R – I > R – Br > R – Cl.

Question xi.
Write reaction to convert ethylamine into methylamine.
Answer:

4. Answer the following.

Question i.
Write the IUPAC names of the following amines :

Answer:

Question ii.
What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).

(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.

Examples :
(i) CH₃ – NH₂ methylamine
(ii) CH₃ – CH₂ – NH₂ ethylamine

(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.

Examples :
(i) C₂H₅ – NH – CH₃ ethylmethylamine
(ii) CH₃ – NH – CH₃ dimethylamine

(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.

Examples :

Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.

(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.

(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.

Question iii.
Write IUPAC names of the following amines.

Answer:

Question iv.
Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :

b. Ethanamine from nitroethane :

c. Ethanamine from Propionamide :

Question v.
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Answer:
Acetylation of amines : The reaction in which the H atom attached to nitrogen in amine is replaced by acetyl group is called acetylation of amines.

(1) Ethylamine on reaction with acetic anhydride forms monoacetyl derivative, N-acetylethylamine.

(2) Diethylamine (a secondary amine) on reaction with acetic anhydride forms a monoacetyl derivative, N-acetyldiethyl amine (or N,N-diethyl acetamide).

(3) Triethylamine does not react with acetic anhydride as it does not have any H atom attached nitrogen atom of amin e

Question vii.
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent?
Answer:
This reaction is useful for the distinction of primary, secondary and tertiary amines.

(i) Primary amine (like ethyl amine) is treated with Hinsberg’s reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.

(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.

(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.

Question viii.
Write reactions to bring about the following conversions :
a. Aniline into p-nitroaniline
b. Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline

(2) Aniline into sulphanilic acid.

Activity :

• Prepare a chart of azodyes, colours and its application.
• Prepare a list of names and structures of N-containing ingredients of diet.

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

Use your brain power! (Textbook Page No 282)

Question 1.
Classify the following amines as simple/mixed; 1°, 2°, 3° and aliphatic or aromatic. (C₂H₅)₂NH, (CH₃)₃N, C₂H₅ – NH – CH₃,

Answer:

(A) Common Names : Rules

1. According to common naming system, the amines are named as alkylamines.
2. The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word ‘amine’.
   Example : CH₃ – NH₂ : methyl-amine
3. The simple {symmetrical) secondary and tertiary amines are written by adding prefix ‘di- (forpresence of two alkyl groups) and ‘tri’- (for presence of three alkyl groups) respectively to the name of alkyl groups.
   Examples: (i) CH₃ – NH – CH₃ dimethylamine, (ii) (C₂H₅)₃ N triethylamine
4. The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word ‘amine’.
   Example : CH₃ – CH₂ – NH – CH₃ ethyhnethylamine

(B) IUPAC names : Rules

1. According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
2. The name of the amine is obtained by replacing the suffix ‘e’ from parent alkane’s name by ‘amine’.
3. The position of the amino group is indicated by the lowest possible locant.
   Example :
   
4. In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
   Example : ClH₅NH – CH₃ N – Methylethanamine
5. A complete name of amine is written as one word.

Try this….. (Textbook Page No 283)

Question 1.
Draw possible structures of all the isomers of C₄H₁₁N. Write their common as well as IUPAC names.
Answer:

Use your brain power! (Textbook Page No 283)

Question 1.
Write chemical equations for

(i) reaction of alc. NH with C₂H₅I.
Answer:

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:

(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer:
In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.

Problem 13.1 : (Textbook Page No 285)

Question 1.
Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution :
Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.

The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.

Use your brain power! (Textbook Page No 285)

Identify ‘A’ and ‘B’ in the following conversions.

Answer:

Use your brain power! (Textbook Page No 286)

Question 1.
Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine

(2) Benzamide to aniline

(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine

(4) Benzamide to benzylamine

Use your brain power! (Textbook Page No 287)

Question 1.
Arrange the following :
(1) In decreasing order of the boiling point C₂H₅ – OH, C₂H₅ – NH₂, (CH₃)₂ NH
(2) In increasing order of solubility in water: C₂H₅ – NH₂, C₃H₇ – NH₂, C₆H₅ – NH₂
Answer:
(1) Decreasing order of the boiling point : C₂H₅ — OH, C₂H₅ — NH₂, (CH₃)₂ NH
(2) Increasing order of solubility in water : C₆H₅NH₂, C₃H₇ — NH₂, C₂H₅ — NH₂

Use your brain power! (Textbook Page No 288)

Question 1.
Refer to pK_b values and answer which compound from the following pairs is the stronger base?
(1) CH₃ – NH₂ and (CH₃)₂ NH
(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(3) NH₃ and (CH₃)₂ CH – NH₂
Answer:
(1) CH₃ -NH₂ and (CH₃)₂ NH
(CH₃)₂ NH is a stronger base

(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(C₂H₅)₂ NH is a stronger base

(3) NH₃ and (CH₃)₂ CHNH₂
(CH₃)₂ CHNH₂ is a stronger base

Use your brain power! (Textbook Page No 290)

Question 1.
Arrange the following amines in decreasing order of their basic strength :
NH₃, CH₃ – NH₂, (CH₃)₂ NH, C₆H₅NH₂
Answer:
Decreasing order of basic strength :
(CH₃)₂NH, CH₃ -NH₂, NH₃, C₆H₅NH₂

Use your brain power! (Textbook Page No 291)

Question 1.

Answer:

Use your brain power! (Textbook Page No 291)

Question 1.
Complete the following reaction :

Answer:

Use your brain power! (Textbook Page No 292)

Answer:

Use your brain power! (Textbook Page No 292)

Question 1.
Write the carbylamine reaction by using aniline as starting material.
Answer:

Can you tell? (Textbook Page No 292)

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H – O – N = O
(2) Nitrous acid cannot be stored in bottle.

Use your brain power! (Textbook Page No 294)

Question 1.
How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg’s test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide

(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonamide.

(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH

Problem 13.1 : (Textbook Page No 295)

Question 1.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution :
NH₂ – group in aniline is highly ring activating and o – /p – directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with Br₂ it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of – NH₂ group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.

Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original – NH₂ group is regenerated. The protection of – NH₂ group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.

Use your brain power! (Textbook Page No 296)

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel – Craft’s reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr’s reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for the electrophilic attack.

Can you tell? (Textbook Page No 294)

(1) Do tertiary amines have ‘H’ bonded to ‘N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
(1) Tertiary amines do not have ‘H’ bonded to ‘N’.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to the nitrogen atom of amine.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

1. Choose the most correct option.

Question i.
In the following resonating structures A and B, the number of unshared electrons in valence shell present on oxygen respectively are

Answer:
c. 4, 6

Question ii.
In the Wolf -Kishner reduction, alkyl aryl ketones are reduced to alkylbenzenes. During this change, ketones are first converted into
a. acids
b. alcohols
c. hydrazones
d. alkenes
Answer:
c. hydrazones

Question iii.
Aldol condensation is
a. electrophilic substitution reaction
b. nucleophilic substitution reaction
c. elimination reaction
d. addition – elimination reaction
Answer:
d. addition-elimination reaction

Question iv.
Which one of the following has the lowest acidity?


Answer:
(c)

Question v.
Diborane reduces
a. ester group
b. nitro group
c. halo group
d. acid group
Answer:
d. acid group

Question vi.
Benzaldehyde does NOT show positive test with
a. Schiff reagent
b. Tollens’ ragent
c. Sodium bisulphite solution
d. Fehling solution
Answer:
d. Fehling solution

2. Answer the following in one sentence

Question i.
What are aromatic ketones?
Answer:
The compounds in which group is attached to either two aryl groups or one aryl and one alkyl group are called aromatic ketones.

For example :

Question ii.
Is phenylacetic acid an aromatic carboxylic acid?
Answer:
Phenylacetic acid is not an aromatic carboxylic acid.

Question iii.
Write reaction showing conversion of ethanenitrile into ethanol.
Answer:

Question iv.
Predict the product of the following reaction:
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOCH}_{3} \frac{\mathrm{i} \cdot \mathrm{AlH}(\mathrm{i}-\mathrm{Bu})_{2}}{\text { ii. } \mathrm{H}_{3} \mathrm{O}^{\oplus}}{\longrightarrow} ?\)
Answer:

Question v.
Name the product obtained by reacting toluene with carbon monoxide and hydrogen chloride in presence of anhydrous aluminium chloride.
Answer:

Question vi.
Write reaction showing conversion of Benzonitrile into benzoic acid.
Answer:

Question vii.
Name the product obtained by the oxidation of 1,2,3,4-tetrahydronaphthalene with acidified potassium permanganate.
Answer:

Question viii.
What is formalin?
Answer:
The aqueous solution of formaldehyde (40%) is known as formalin.

Question ix.
Arrange the following compounds in the increasing order of their boiling points : Formaldehyde, ethane, methyl alcohol.
Answer:
Ethane, formaldehyde, methyl alcohol.

Question x.
Acetic acid is prepared from methyl magnesium bromide and dry ice in presence of dry ether. Name the compound which serves not only reagent but also as cooling agent in the reaction.
Answer:
The cooling agent used in the above reaction is dry ice (O = C = O).

3. Answer in brief.

Question i.
Observe the following equation of reaction of Tollens’ reagent with aldehyde. How do we know that a redox reaction has taken place. Explain.
\(\begin{array}{r}
\mathrm{R}-\mathrm{CHO}+2 \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}+\mathrm{OH}^{-} \stackrel{\Delta}{\longrightarrow} \\
\mathrm{R}-\mathrm{COO}^{-}+2 \mathrm{Ag} \downarrow+4 \mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}
\end{array}\)
Answer:
Tollen’s reagent oxidises acetaldehyde to acetic acid (carboxylate ion) and Ag in Tollen’s reagent complex are reduced to silver. In this reaction, oxidation and reduction takes place simultaneously hence, it is a redox reaction.

Question ii.
Formic acid is stronger than acetic acid. Explain.
Answer:

In acetic acid, the methyl group is an electron-donating group. The acetate ion formed gets destabilized due to the electron releasing effect of methyl group ( +1 effect) which is higher than that of H-atom in the corresponding formic acid. As a result, acetic acid dissociates to a lesser extent. Thus decreasing the acidity of acetic acid.

Formic acid having lower pK_a value than acetic acid. Hence, formic acid is a stronger acid than acetic acid.

Question iii.
What is the action of hydrazine on cyclopentanone in presence of —. KOH in ethylene glycol?
Answer:

Question iv.
Write reaction showing conversion of Acetaldehyde into acetaldehyde dimethyl acetal.
Answer:

Question v.
Aldehydes are more reactive toward nucleophilic addition reactions than ketones. Explain.
Answer:
The reactivity of aldehydes and keones is due to the polarity of carbonyl group which results in electrophilicity of carbon. The reactivity is further explained on the basis of electronic effect and steric effects.

(1) Influence of electronic effects: A ketone has two electron-donating aJ.kyl groups ( + I effect) bonded to carbonyl carbon which are responsible for decreasing its positive polarity and electrophilicity. In contrast. aldehydes have only electron-donating group bonded to carbonyl carbon. This shows aldehydes are more electrophilic than ketones.

(2) Steric effects : Two bulky alkyl groups in ketone come in the way of the incoming nucleophile. This is called steric hindrance to nucleophilic attack.

On the other hand. nucleophile can easily attack the carbonyl carbon in aldehyde because has one alkyl group and is less crowded or sterically less hindered.

Hence aldehydes are more reactive and can easily be attacked by nucleophiles.

Question vi.
Write reaction showing the action of the following reagent on propane nitrile
a. Dilute NaOH
b. Dilute HCl ?
Answer:
(1) Action of dil NaOH:

(2) Action of dil HCl:

Question vi.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Answer:
The increasing order of acidity will be 1 <3 <2. Acidity depends on mainly two factors : (1) ease of proton release (2) stability of conjugate base formed. In example (3) the ether O exerts a I effect and is closer to COOH group than in 2 (1 effect diminishes). Also the conjugate base formed will be stabilized by the same – I effect by delocalization of charge.

4. Answer the following

Question i.
Write a note on
a. Cannizaro reaction
b. Stephen reaction.
Answer:
(1) The carbon atom adjacent to carbonyl carbon atom is called a-carbon atom (α – C) and the hydrogen atom attached to a-carbon atom is called α-hydrogen atom (α – H).

(2) The α-hydrogen of aldehydes and ketones is acidic in nature due to (i) the strong-I effect of carbonyl group (ii) resonance stabilization of the carbanion.

(3) Aldol condensation reaction is characteristic reaction of aldehydes and ketones containing active α-hydrogen atom.

(4) When aldehydes or ketones containing α – H atoms are warmed with a dilute base or dilute acid, two molecules of them undergo self condensation to give β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol) respectively. The reaction is known as Aldol addition Reaction.

(5) In aldol condensation, the product is formed by the nucleophilic addition of α-carbon atom of a second molecule which gets attached to carbonyl carbon atom of the first molecule and α-hydrogen atom of the second molecule gets attached to carbonyl oxygen atom of the first molecule forming (- OH) group to give β-hydroxy aldehyde or ketone.

(6) This is a reversible reaction, establishing an equilibrium favouring aldol formation to a greater extent than ketol formation.

(7) For aldehyde :

Acetaldol on heating undergoes subsequent elimination of water giving rise to α, β unsaturated aldehyde.

The overall reaction is called aldol condensation. It is a nucleophilic addition-elimination reaction. For ketone :

Diacetone alcohol on dehydration by heating forms α, β unsaturated ketone.

Question ii.
What is the action of the following reagents on toluene ?
a. Alkaline KMnO₄, dil. HCl and heat
b. CrO₂Cl₂ in CS₂
c. Acetyl chloride in presence of anhydrous AlCl₃.
Answer:
(1) Action of alkaline KMnO₄ : When toluene is heated with alkaline KMnO₄. (methyl group gets oxidised to earboxy lic group) benzoic acid is obtained

(2) Action of CrO₂Cl₂ in C₂:
Answer:
When Loluenc is ircated with soluion of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde.

(3) Action of acetyl chloride in presence of anhyd. AlCl₃.
Answer:
When toluene is treated with acetyl chloride in presence of anhydrous AlCl₃ 4-methyl acetophenone is obtained.

Question iii.
Write the IUPAC names of the following structures :

Question iv.
Write reaction showing conversion of p- bromoisopropyl benzene into p-Isopropyl benzoic acid (3 steps).
Answer:

Question v.
Write reaction showing aldol condensation of cyclohexanone.
Answer:

Activity :
Draw and complete the following reaction scheme which starts with acetaldehyde. In each empty box, write the structural formula of the organic compound that would be formed.

12th Chemistry Digest Chapter 12 Aldehydes, Ketones and Carboxylic Acids Intext Questions and Answers

Use your brain power! (Textbook Page No 254)

Question 1.
Classify the following as aliphatic and aromatic aldehydes.

Answer:

Use your brain power! (Textbook Page No 255)

Question 1.
Classify the following as simple and mixed ketones. Benzoplienone, acetoneq hutanoneq acetophenone.

Compoun
Benzophenone	……………………………………………..
Acetone	……………………………………………..
Butanone	……………………………………………..
Acetophenone	……………………………………………..

Answer:

Compound
Benzophenone	Simple ketone
Acetone	Simple ketone
Butanone	Mixed ketone
Acetophenone	Mixed ketone

Use your brain power! (Textbook Page No 264)

Write IUPAC names for the following compounds.

Try this ….. (Textbook Page No 260)

Question 1.
Draw structures for the following :
(1) 2-Methylpentanal
(2) Hexan-2-one
Answer:

Can you tell? (Textbook Page No 260)

Question 1.
Which is the reagent that oxidizes primary alcohols to only aldehydes and does not oxidize aldehydes further into carboxylic acid ?
Answer:
The reagent that is used to make only aldehydes is-heated Cu at 573 K.

Use your brain power! (Textbook Page No 261)

Question 1.
Write the structure of the product formed on Rosenmund reduction of ethanoyl chloride and benzoyl chloride.
Answer:

Can you think? (Textbook Page No 261)

Question 1.
What is the alcohol formed when benzoyl chloride is reduced with pure palladium as the catalyst ?
Answer:

Use your brain power! (Textbook Page No 262)

Question 1.
Name the compounds which are used for the preparation of benzophenone by Friedel-Crafts acylation reaction. Draw their structures.
Answer:
The compounds which are used in preparation of benzophenone by Friedel – Crafts reaction are :

Use your brain power! (Textbook Page No 263)

Identify the reagents necessary to achieve each of the following transformations:

Answer:

Use your brain power! (TextBook Page No 264)

Predict the products (name and structure) in the following reactions.

Answer:

Problem 12.1 : (Textbook Page No 276)

Question 1.
Alcohols (R – OH), phenols (Ar – OH) and carboxylic acids (R – COOH) can undergo ionization of O – H bond to give away proton H⁺; yet they have different pK_a values, which are 16, 10 and 4.5 respectively. Explain.
Solution :
pK_a value is indicative of acid strength. Lower of pK_a value stronger the acid. Alcohols, phenols and carboxylic acids, all involve ionization of an O – H bond. But their different pK_a values indicate that their acid strengths are different. This is because the resulting conjugate bases are stabilized to different extents.

As the conjugate base of carboxylic acid is best stabilized, among the three, carboxylic acids are strongest and have the lowest pKa value. As conjugate base of alcohols is destabilized, alcohols are weakest acids and have highest pIQ value. As conjugate base of phenols is moderately stabilized, phenols are moderately acidic and have intermediate pBQ value.

Try this….. (Textbook Page No 277)

Question 1.
Compare the following two conjugate bases and answer.

(1) Indicate the inductive effects of CH₃ – group in (a) and Cl-group in (b) by putting arrowheads in the middle of appropriate covalent bonds.
(2) Which species is stabilized by inductive effect, (a) or (b) ?
(3) Which species is destabilized by inductive effect, (a) or (b) ?
Answer:

(2) The monochloroacetate ion formed gets stabilised due to electron-withdrawing of Cl atom (- I effect).
(3) The acetate ion formed gets destabilised due to electron releasing effect of methyl group

Use your brain power! (Textbook Page No 277)

Question 1.
(1) Compare the pK_a values and arrange the following in an increasing order of acid strength. CI₃CCOOH, ClCH₂COOH, CH₃COOH, Cl₂CHCOOH
(2) Draw structures of conjugate bases of monochloroacetic acid and dichloroacetic acid. Which one is more stabilized by – 1 effect?
Answer:

The dichloroacetate ion formed gets stabilised due electron-withdrawing effect of two chlorine atoms. (- 1 effect)

Try this….. (Textbook Page No 277)

Question 1.
Arrange the following acids in order of their decreasing acidity.

Answer:
Acidity in the decreasing order

Try this ….. (Textbook Page No 267)

Question 1.
Draw the structure of propanone and indicate its polarity.
Answer:

Can you tell? (Textbook Page No 268)

Question 1.
Simple hydrocarbons, ethers, ketones and alcohols do not get oxidized by Tollen’s reagent. Explain, Why?
Answer:
(1) Due to the presence of hydrogen atom in aldehyde group , an aldehyde is oxidised to carboxylic acid which is not possible in case of ethers, ketones, alcohols and hydrocarbons.
(2) In ketones, carbonyl atom is attached to C-atom, hence C – C bond in can’t be broken easily.
(3) H atom attached to carbonyl carbon can be oxidised to – OH group giving carboxylic group Therefore, aldehyde reduces Tollen’s reagent, whereas simple hydrocarbons, ethers, ketones and alcohols do not reduce Tollen’s reagent.

Use your brain power! (Textbook Page No 269)

Question 1.
Sodium bisulfite is sodium salt of sulfurous acid, write down its detailed bond structure.
Answer:

Bond structure of sodium bisulfite

Use your brain power! (Textbook Page No 270)

Predict the product of the following reaction:

Answer:

Use your brain power! (Textbook Page No 271)

Question 1.
Draw the structures of
(1) The semicarbazone of cyclohexanone
(2) The imine formed in the reaction between 2-methylhexanal and ethyl amine
(3) 2, 4-dinitrophenyl hydrazone of acetaldehyde.
Answer:
(1) The semi carbazone of cyclohexanone.

(2) The imine formed between 2-methyl hexanal and ethyl amine.

(3) 2, 4-dinitrophenylhydrazone of acetaldehyde.

Try this….. (Textbook Page No 272)

Question 1.
Write chemical reactions taking place when propan-2-ol is treated with iodine and sodium hydroxide.
Answer:

Question 2.
When acetaldehyde Is treated with dilute NaOH, the following reaction is observed.

(1) What are the functional groups in the product?
(2) Can another product be formed during the same reaction? (Deduce the answer by doing atomic audit of reactant and product)
(3) Is this an addition reaction or condensation reaction?
Answer:
(1) There arc two functiona’ groups in the product: -CRO and -OH
(2) No other product can be formed in the same reaction.
(3) This is an addition reaction.

Use your brain power! (Textbook Page No 273)

Question 1.
Observe the following reaction.
Answer:

Question 2.
Will this reaction give a mixture of products like a cross aldol reaction ?
Answer:
No, since benzaldehyde, does not have a-hydrogen atom, it will not undergo self aldol condensation.

Use your brain power! (Textbook Page No 274)

Question 1.
Can isobutyraldehyde undergo a Cannizzaro reaction? Explain.
Answer:
Since isobutyraldehydecontains a-carbon atom, it cannot undergo Cannizzaro reaction.

Can you tell? (Textbook Page No 279)

What is the term used for elimination of water molecule ?
Answer:
Dehydration.

Use your brain power! (Textbook Page No 278)

Question 1.
Fill in the blanks and rewrite the balanced equations.

Answer: