Category: Class 7

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 42 Answers Solutions Chapter 11 Circle.

Circle Class 7 Practice Set 42 Answers Solutions Chapter 11

Question 1.
Complete the table below:

Sr. No	Radius (r)	Diameter (d)	Circumference (c)
i.	7 cm
ii.		28 cm
iii.			616 cm
iv.			72.6 cm

Solution:
i. Radius (r) = 7 cm
Diameter (d) = 2r
= 2 x 7 = 14 cm
Circumference (c) = πd
= \(\frac { 22 }{ 7 }\) x 14
= 44 cm

ii. Diameter (d) = 28 cm
Radius (r) = \(\frac{d}{2}=\frac{28}{2}\) = 14 cm
Circumference (c) = πd
= \(\frac { 22 }{ 7 }\) x 28
= 88 cm

iii. Circumference (c) = 616 cm
∴ πd = 616
∴ \(\frac { 22 }{ 7 }\) x d = 616
∴ d = 616 x \(\frac { 7 }{ 22 }\)
∴ d = 196 cm
∴ Diameter (d) = 196 cm
Radius (r) = \(\frac{\mathrm{d}}{2}=\frac{196}{2}\) = 98 cm

iv. Circumference (c) = 72.6 cm
∴ πd = 72.6
\(\frac { 22 }{ 7 }\) x d = 72.6
∴ \(d=72.6 \times \frac{7}{22}=\frac{726}{10} \times \frac{7}{22}=\frac{33 \times 7}{10}\)
∴ d = 23.1 cm
∴ Diameter (d) = 23.1 cm
Radius (r) = \(\frac{\mathrm{d}}{2}=\frac{23.1}{2}\)
= 11.55 cm

Sr. No	Radius (r)	Diameter (d)	Circumference (c)
i.	7 cm	14 cm	44 cm
ii.	14 cm	28 cm	88 cm
iii.	98 cm	196 cm	616 cm
iv.	11.55 cm	23.1 cm	72.6 cm

Question 2.
If the circumference of a circle is 176 cm, find its radius.
Solution:
Circumference (c) = 176 cm
∴ 2πr = 176
∴ 2 x \(\frac { 22 }{ 7 }\) x r = 176
∴ \(\frac { 44 }{ 7 }\) x r = 176
∴ r = 176 x \(\frac { 7 }{ 44 }\) = 28 cm
∴ The radius of the circle is 28 cm.

Question 3.
The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre?
Solution:
Radius of the circular garden (r) = 56 m
∴ Circumference of the circular garden (c) = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 56
= 352 m
∴ Length of the wire required to put 1-round fence = Circumference
∴ Length of wire required to put a 4-round fence = 4 x Circumference
= 4 x 352
= 1408 m
∴ Cost of wire per meter = Rs 40
∴ Total cost = length of wire required x cost of the wire
= 1408 x 40
= Rs 56320
∴ The cost to put a 4-round fence around the garden is Rs 56320.

Question 4.
The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km?
Solution:
Diameter of the wheel of the bullock cart (d) = 1.4 m
Circumference of the wheel of the bullock cart (c) = πd
\(=\frac{22}{7} \times 1.4=\frac{22}{7} \times \frac{14}{10}=\frac{44}{10}=4.4 \mathrm{m}\)
Distance covered in 1 rotation = Circumference of the wheel
= 4.4 m

∴ The wheel of the bullock cart will complete 250 rotations as the cart travels 1.1 km.

Maharashtra Board Class 7 Maths Chapter 11 Circle Practice Set 42 Intext Questions and Activities

Question 1.
Identify the radii, chords and diameters in the circle alongside and write their names in the table below: (Textbook pg. no. 75)

i. Radii
ii. Chords
iii. Diameters

Solution:
i. OA, OB, OC, OF
ii. EC, AD, AB, FC
iii. AB, FC

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 33 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 33 Answers Solutions Chapter 8

Question 1.
Add:
i. 9p + 16q; 13p + 2q
ii. 2a + 6b + 8c; 16a + 13c + 18b
iii. 13x² – 12y²; 6x² – 8y²
iv. 17a²b² + 16c; 28c – 28a²b²
v. 3y² – 10y + 16; 2y – 7
vi. – 3y² + 10y – 16; 7y² + 8
Solution:
i. (9p + 16q) + (13p + 2q)
= (9p + 13p) + (16q + 2q)
= 22p + 18q

ii. (2a + 6b + 8c) + (16a + 13c + 18b)
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c

iii. (13x² – 12y²) + (6x² – 8y²)
= (13x² + 6x²) + [(-12y²) + (-8y²)]
= 19x² + (-20y²)
= 19x² – 20y²

iv. (17a²b² + 16c) + (28c – 28a²b²)
= [17a²b² + (-28a²b²)] + (16c + 28c)
= -11a²b² + 44c

v. (3y² – 10y + 16) + (2y – 7)
= 3y² + (-10y + 2y) + (16 – 7)
= 3y² – 8y + 9

vi. (-3y² + 10y – 16) + (7y² + 8)
= (-3y² + 7y²) + (10y) + (-16 + 8)
= 4y² + 10y – 8

Maharashtra Board Class 7 Maths Chapter 8 Algebraic Expressions and Operations on them Practice Set 33 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 57)

1. 3x + 4y = How many?
2. 3 guavas + 4 mangoes = 7 guavas.
3. 7m – 2n = 5m.

Solution:

1. 3x and 4y are unlike terms. Hence, they cannot be added, further to get a single term.
2. No. Guava and mango are different fruits. Hence, 3 guavas + 4 mangoes & 7 guavas.
3. No. 7m and 2n are unlike terms. Hence, 7m – 2n ≠ 5m.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 41 Answers Solutions Chapter 10 Bank and Simple Interest.

Bank and Simple Interest Class 7 Practice Set 41 Answers Solutions Chapter 10

Question 1.
If the interest on Rs 1700 is Rs 340 for 2 years, the rate of interest must be__.
(A) 12%
(B) 15%
(C) 4%
(D) 10%
Solution:
(D) 10%

Hint:
∴ \(\text { Total interest }=\frac{P \times R \times T}{100}\)
∴ \(340=\frac{1700 \times R \times 2}{100}\)
∴ R = 10%

Question 2.
If the interest on Rs 3000 is Rs 600 at a certain rate for a certain number of years, what would the interest be on Rs 1500 under the same conditions?
(A) Rs 300
(B) Rs 1000
(C) Rs 700
(D) Rs 500
Solution:
(A) Rs 300

Hint:
The interest on Rs 3000 at certain rate of interest is Rs 600.
Let us suppose the interest on Rs 1500 at the same rate is x.
∴ \(\frac{600}{3000}=\frac{x}{1500}\)
∴ x = Rs 300

Question 3.
Javed deposited Rs 12000 at 9 p.c.p.a in a bank for some years, and withdrew his interest every year. At the end of the period, he had received altogether Rs 17,400. For how many years had he deposited his money?
Solution:
Here, P = Rs 12000, R = 9 p.c.p.a and amount = Rs 17400
Amount = Principal + Interest
∴17400 = 12000 + Interest
∴Interest = 17400 – 12000 = Rs 5400
∴ \(\text { Total interest }=\frac{P \times R \times T}{100}\)
\(5400=\frac{12000 \times 9 \times \mathrm{T}}{100}\)
∴ \(\frac{5400 \times 100}{12000 \times 9}=\mathrm{T}\)
∴ T = 5 years
∴ Javed had deposited the amount for 5 years.

Question 4.
Lataben borrowed some money from a bank at a rate of 10 p.c.p.a interest for \(2\frac { 1 }{ 2 }\) years to start a cottage industry. If she paid Rs 10250 as total interest, how much money had she borrowed?
Solution:
Here, R = 10 p.c.p.a, T = 2.5 years, I = Rs 10250

∴ P = Rs 41000
∴ Lataben had borrowed an amount of Rs 41000 from the bank.

Question 5.
Fill in the blanks in the table.

	Principal	Rate of interest (p.c.p.a.)	Time	Interest	Amount
i.	Rs 4200	7%	3 years
ii.		6%	4 years	Rs 1200
iii.	Rs 8000	5%		Rs 800
iv.		5%		Rs 6000	Rs 18000
v.		\(2\frac { 1 }{ 2 }\) % 2	5 years	Rs 2400

Solution:
i. \(\text { Total interest }=\frac{P \times R \times T}{100}\)
= \(\frac{4200 \times 7 \times 3}{100}\)
= Rs 882
Amount = Principal + interest
= 4200 + 882
= Rs 5082

ii. \(\text { Total interest }=\frac{P \times R \times T}{100}\)
∴ \(1200=\frac{\mathrm{P} \times 6 \times 4}{100}\)
∴ \(\frac{1200 \times 100}{6 \times 4}=\mathrm{P}\)
∴ P = Rs 5000
Amount = Principal + interest
= 5000 + 1200
= Rs 6200

iii. \(\text { Total interest }=\frac{P \times R \times T}{100}\)
∴ \(800=\frac{8000 \times 5 \times \mathrm{T}}{100}\)
∴ \(\frac{800 \times 100}{8000 \times 5}=\mathrm{T}\)
∴ T = 2 years
Amount = Principal + interest
= 8000 + 800
= Rs 8800

iv. Amount = Principal + interest
∴ 18000 = Principal + 6000
∴ Principal = Rs 12000
\(\text { Total interest }=\frac{P \times R \times T}{100}\)
∴ \(6000=\frac{12000 \times 5 \times \mathrm{T}}{100}\)
∴ \(\frac{6000 \times 100}{12000 \times 5}=\mathrm{T}\)
∴ T = 10 years

v. R = \(2\frac { 1 }{ 2 }\) % = 2.5 %
∴ \(\text { Total interest }=\frac{P \times R \times T}{100}\)
∴ \(2400=\frac{\mathrm{P} \times 2.5 \times 5}{100}\)
∴ \(2400=\frac{P \times 25 \times 5}{100 \times 10}\)
∴ \(\frac{2400 \times 10 \times 100}{25 \times 5}=P\)
∴ P = Rs 19200
Amount = Principal + interest
= 19200 + 2400
= Rs 21600

	Principal	Rate of interest (p.c.p.a.)	Time	Interest	Amount
i.	Rs 4200	7%	3 years	Rs 882	Rs 5082
ii.	Rs 5000	6%	4 years	Rs 1200	Rs 6200
iii.	Rs 8000	5%	2 years	Rs 800	Rs 8800
iv.	Rs 12000	5%	10 years	Rs 6000	Rs 18000
v.	Rs 19200	\(2\frac { 1 }{ 2 }\) % 2	5 years	Rs 2400	Rs 21600

Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 41 Intext Questions and Activities

Question 1.
Ask an adult in your house to show you a passbook and explain the entries made in it. (Textbook pg. no. 70)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)

Question 2.
Visit different banks and find out the rates of the interest they give for different types of accounts. (Textbook pg. no. 74)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)

Question 3.
With the help of your teachers, start a Savings Bank in your school and open an account in it to save up some money. (Textbook pg. no. 74)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 38 Answers Solutions Chapter 9 Direct Proportion and Inverse Proportion.

Direct Proportion and Inverse Proportion Class 7 Practice Set 38 Answers Solutions Chapter 9

Question 1.
Five workers take 12 days to weed a field. How many days would 6 workers take? How many would 15 take?
Solution:
Let 6 workers take x days and 15 workers take y days to weed the field.
The number of workers and the time required to weed the field are in inverse proportion.
∴ 6 × x = 5 × 12
∴ \(x=\frac{5 \times 12}{6}\)
∴ x = 10 days
Also, 15 × y = 5 × 12
∴ \(y=\frac{5 \times 12}{15}\)
= 4 days
∴ 6 workers will take 10 days and 15 workers will take 4 days to weed the field.

Question 2.
Mohanrao took 10 days to finish a book, reading 40 pages every day. How many pages must he read in a day to finish it in 8 days?
Solution:
Let Mohanrao read x pages every day to finish the book in 8 days.
The number of pages read per day and the days required to finish the book are in inverse proportion.
∴ 8 × x = 10 × 40
∴ \(x=\frac{10 \times 40}{8}\)
= 50
∴ Mohanrao will have to read 50 pages every day to finish the book in 8 days.

Question 3.
Mary cycles at 6 km per hour. How long will she take to reach her Aunt’s house which is 12 km away? If she cycles at a speed of 4 km/hr, how long would she take?
Solution:
Speed of the cycle = 6 km / hr
Distance travelled to reach her Aunt’s house = 12 km
∴ \(\text { Time required }=\frac{\text { Distance travelled }}{\text { Speed }}\)
= \(\frac { 12 }{ 6 }\)
= 2 hours
Let the time required when the speed of the cycle is 4 km/hr be x hours.
The speed of the cycle and the time required to travel the same distance are in inverse proportion.
∴ 4 × x = 6 × 2
∴ \(x=\frac{6 \times 2}{4}\) = 3 hours
∴ Mary will require 2 hours if she is cycling at 6 km/hr and 3 hours if she is cycling at 4 km/hr to reach her Aunt’s house.

Question 4.
The stock of grain in a government warehouse lasts 30 days for 4000 people. How many days will it last for 6000 people?
Solution:
Let the stock of grain last for x days for 6000 people.
The number of people and the days for which stock will last are in inverse proportion.
∴ 6000 × x = 4000 × 30
∴ \(x=\frac{4000 \times 30}{6000}=20\)
∴ The stock of grain will last for 20 days for 6000 people.

Maharashtra Board Class 7 Maths Chapter 9 Direct Proportion and Inverse Proportion Practice Set 38 Intext Questions and Activities

Question 1.
Students of a certain school went for a picnic to a farm by bus. Here are some of their experiences. Say whether the quantities in each are in direct or in inverse proportion.
(Textbook pg. no. 65 and 66)
i. Each student paid Rs 60 for the expenses.
As there were 45 students,___rupees were collected.
Had there been 50 students,___rupees would have been collected.
The number of students and money collected are in___proportion.

ii. The sweets shop near the school gave 90 laddoos for the picnic.
If 45 students go for the picnic, each will get___laddoos.
If 30 students go for the picnic, each will get___laddoos.
The number of students and that of laddoos each one gets are in___proportion.

iii. The farm is 120 km away from the school.
The bus went to the farm at a speed of 40 km per hour and took___hours.
On the return trip, the speed was 60 km per hour. Therefore, it took___hours.
The speed of the bus and the time it takes are in___proportion.

iv. The farmer picked 180 bors from his trees.
He gave them equally to 45 students. Each student got___bors.
Had there been 60 students, each would have got___bors.
The number of students and the number of bors each one gets are in___proportion.
Solution:
i. Rs 2700, Rs 3000, direct
ii. 2,3, inverse
iii. 3,2, inverse
iv. 4,3, inverse

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 39 Answers Solutions Chapter 9 Direct Proportion and Inverse Proportion.

Direct Proportion and Inverse Proportion Class 7 Practice Set 39 Answers Solutions Chapter 9

Question 1.
Suresh and Ramesh together invested Rs 144000 in the ratio 4 : 5 and bought a plot of land. After some years they sold it at a profit of 20%. What is the profit each of them got?
Solution:
Total investment = Rs 144000
Profit earned = 20%
∴ Total profit = 20% of 144000 = \(\frac{20}{100} \times 144000\) = Rs 28800
Proportion of investment of Suresh and Ramesh = 4:5
Let the profit of Suresh be Rs 4x and that of Ramesh be Rs 5x.
4x + 5x = 28800
∴ 9x = 28800
∴ \(x=\frac { 28800 }{ 9 }\)
= 3200
∴ Suresh’s profit = 4x = 4 × 3200 = Rs 12800
Ramesh’s profit = 5x = 5 × 3200 = Rs 16000
∴ The profit earned by Suresh and Ramesh are Rs 12800 and Rs 16000 respectively.

Question 2.
Virat and Samrat together invested Rs 50000 and Rs 120000 to start a business. They suffered a loss of 20%. How much loss did each of them incur?
Solution:
Total investment = Rs 50000 + Rs 120000 = Rs 170000
Loss incurred = 20%
∴ Total loss = 20% of 170000 = \(\frac{20}{100} \times 170000\) = Rs 34000
Proportion of investment = 50000 : 120000
= 5 : 12 …. (Dividingby 10000)
Let the loss incurred by Virat be Rs 5x and that by Samrat be Rs 12x.
5x + 12x = 34000
∴ 17x = 34000
∴ \(x=\frac { 34000 }{ 17 }=2000\)
∴ Virat’s loss = 5x = 5 × 2000 = Rs 10000
Samrat’s loss = 12x = 12 × 2000 = Rs 24000
∴ The loss incurred by Virat and Samrat are Rs 10000 and Rs 24000 respectively.

Question 3.
Shweta, Piyush and Nachiket together invested Rs 80000 and started a business of selling sheets and towels from Solapur. Shweta’s share of the capital was Rs 30000 and Piyush’s Rs 12000. At the end of the year they had made a profit of 24%. What was Nachiket’s investment and what was his share of the profit?
Solution:
Total investment = Rs 80000
Nachiket’s investment = Total investment – (Shweta’s investment + Piyush’s investment)
= 80000 – (30000+ 12000)
= 80000 – 42000 = Rs 38000
Profit earned = 24%
∴ Total profit = 24% of 80000 = \(\frac { 24 }{ 100 }\) x 80000 = Rs 19200
Proportion of investment = 30000 : 12000 : 38000
= 15 : 6 : 19 …. (Dividing by 2000)
Let the profit of Shweta, Piyush and Nachiket be Rs 15x, Rs 6x and Rs 19x respectively.
15x + 6x + 19x = 19200
∴ 40x = 19200
∴ \(x=\frac { 19200 }{ 40 }=480\)
∴ Nachiket’s profit = 19x = 19 × 480 = Rs 9120
∴ Nachiket’s investment is Rs 38000 and his profit is Rs 9120.

Question 4.
A and B shared a profit of Rs 24500 in the proportion 3 : 7. Each of them gave 2% of his share of the profit to the Soldiers’ Welfare Fund. What was the actual amount given to the Fund by each of them?
Solution:
Proportion of share = 3:7
Let the profits of A and B be Rs 3x and Rs 7x respectively.
3x + 7x = 24500
∴ 10x = 24500
∴ \(x=\frac { 24500 }{ 10 }=2450\)
Profit earned by A = 3x = 3 × 2450 = Rs 7350
Amount given by A = 2% of his profit
= \(\frac { 2 }{ 100 }\) × 7350 = Rs 147
Profit earned by B = 7x = 7 × 2450 = Rs 17150
Amount given by B = 2% of his profit
= \(\frac { 2 }{ 100 }\) × 17150 = Rs 343
∴ The amount given by A and B to the Soldiers’ Welfare Fund are Rs 147 and Rs 343 respectively.

Question 5.
Jaya, Seema, Nikhil and Neelesh put in altogether Rs 360000 to form a partnership, with their investments being in the proportion 3 : 4 : 7 : 6. What was Jaya’s actual share in the capital? They made a profit of 12%. How much profit did Nikhil make?
Solution:
Total investment = Rs 360000
Profit earned = 12%
∴ Total profit = 12% of 360000
= \(\frac{12}{100} \times 360000\) = Rs 43200
Proportion of investment = 3 : 4 : 7 : 6
Let the investment of Jaya, Seema, Nikhil and Neelesh be Rs 3x, Rs 4x, Rs 7x and Rs 6x respectively.
3x + 4x + 7x + 6x = 360000
∴ 20x = 360000
∴ \(x=\frac { 360000 }{ 20 }\)
= 18000
∴ Jaya’s investment = 3x = 3 x 18000 = Rs 54000
Also, profit made by them is Rs 43200
∴ 3x + 4x + 7x + 6x = 43200
∴ 20x = 43200
∴ \(x=\frac { 43200 }{ 20 }\)
= 2160
∴ Nikhil’s profit = 7x = 7 x 2160 = Rs 15120
∴ Jaya’s share in the capital was Rs 54000 and the profit made by Nikhil was Rs 15120.

Maharashtra Board Class 7 Maths Chapter 9 Direct Proportion and Inverse Proportion Practice Set 39 Intext Questions and Activities

Question 1.
Saritaben, Ayesha and Meenakshi started a business by investing Rs 2400, Rs 5200 and Rs 3400. They made a profit of 50%. If they reinvested all their profit by adding it to the capital, find out the proportions of their shares in the capital during the following year. (Textbook pg. no. 67)
Solution:
Total investment = Rs 2400 + Rs 5200 + Rs 3400 = Rs 11000
Total profit = 50% of 11000 = \(\frac{50}{100} \times 11000\) = Rs 5500
Proportion of shares = 2400 : 5200 : 3400
= 12 : 26 : 17 …. (Dividingby 200)
Let the profit of Saritaben, Ayesha and Meenakshi be Rs 12x, Rs 26x and Rs 17x respectively.
12x + 26x + 17x = 5500
∴ 55x = 5500
∴ x = 100
∴ Saritaben’s profit = 12x = 12 × 100 = Rs 1200
Ayesha’s profit = 26x = 26 × 100 = Rs 2600
Meenakshi’s profit = 17x = 17 × 100 = Rs 1700
∴ Saritaben’s new investment = 2400 + 1200 = Rs 3600
Ayesha’s new investment = 5200 + 2600 = Rs 7800
Meenakshi’s new investment = 3400 + 1700 = Rs 5100
∴ New proportion of shares = 3600 : 7800 : 5100
= 12 : 26 : 17 …. (Dividing by 300)
∴ The proportion of the shares in the capital during the following year is 12 : 26 :17

Question 2.
Are the amount of petrol filled in a motorcycle and the distance traveled by it, in direct proportion? (Textbook pg. no. 63)
Solution:
Yes.
If amount of petrol filled in the motorcycle is less, it will travel less distance and if the amount of petrol filled is more, it will travel more distance.
Hence, the amount of petrol filled in the motorcycle and the distance traveled by it are in direct proportion.

Question 3.
Can you give examples from science or everyday life of quantities that vary in direct proportion? (Textbook pg. no. 63)
Solution:

1. Number of chairs and the number of spectators.
2. Quantity (litres) of water and number of vessels required to store the water.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 37 Answers Solutions Chapter 37 Direct Proportion and Inverse Proportion.

Direct Proportion and Inverse Proportion Class 7 Practice Set 37 Answers Solutions Chapter 9

Question 1.
If 7 kg onions cost Rs 140, how much must we pay for 12 kg onions?
Solution:
Let the cost of 12 kg onions be Rs x.
The quantity of onions and their cost are in direct proportion.
∴ \(\frac{7}{140}=\frac{12}{x}\)
∴ 7x = 12 × 140 ….(Multiplying both sides by 140x)
∴ x = \(\frac { 12\times 140 }{ 7 }\)
= 240
We must pay Rs 240 for 12 kg onions.

Question 2.
If Rs 600 buy 15 bunches of feed, how many will Rs 1280 buy?
Solution:
Let the bunches of feed bought for Rs 1280 be x.
The quantity of feed bought and their cost are in direct proportion.
∴ \(\frac{600}{15}=\frac{1280}{x}\)
∴ 600x = 1280 × 15 …. (Multiplying both sides by 15x)
∴ \(x=\frac{1280 \times 15}{600}=32\)
∴ 32 bunches of feed can be bought for Rs 1280.

Question 3.
For 9 cows, 13 kg 500 g of food supplement are required every day. In the same proportion, how much will be needed for 12 cows?
Solution:
Let the food supplement required for 12 cows be x kg.
The quantity of food supplement required and the number of cows are in direct proportion.
∴ \(\frac{13 \mathrm{kg} 500 \mathrm{gram}}{9}=\frac{x \mathrm{kg}}{12}\)
∴ \(\frac{13.5}{9}=\frac{x}{12}\) ….(13 kg 500 gram = 13.5 kg)
∴ 13.5 × 12 = 9x ….(Multiplying both sides by 9 x 12)
∴ \(\frac{13.5 \times 12}{9}=x\)
∴ x = 18
∴ The food supplement required for 12 cows is 18 kg.

Question 4.
The cost of 12 quintals of soyabean is Rs 36,000. How much will 8 quintals cost?
Solution:
Let the cost of 8 quintals of soyabean be Rs x.
The quantity of soyabeans and their cost are in direct proportion.
∴ \(\frac{12}{36000}=\frac{8}{x}\)
∴ 12x = 8 × 36000 ….(Multiplying both sides by 36000x)
∴ \(x=\frac{8 \times 36000}{12}=24000\)
∴ The cost of 8 quintals of soyabean is Rs 24000.

Question 5.
Two mobiles cost Rs 16,000. How much money will be required to buy 13 such mobiles ?
Solution:
Let the cost of 13 mobiles be Rs x.
The quantity of mobiles and their cost are in direct proportion.
∴ \(\frac{2}{16000}=\frac{13}{x}\)
∴ 2x = 13 × 16000 ….(Multiplying both sides by 16000x)
∴ \(x=\frac{13 \times 16000}{2}=104000\)
∴ Rs 104000 will be required to buy 13 mobiles.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 35 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 35 Answers Solutions Chapter 8

Question 1.
Multiply:
i. 16xy × 18xy
ii. 23xy² × 4yz²
iii. (12a + 17b) × 4c
iv. (4x + 5y) × (9x + 7y)
Solution:
i. 16xy × 18xy
= 16 × 18 × xy × xy
= 288x²y²

ii. 23xy² × 4yz²
= 23 × 4 × xy² × yz²
= 92xy³z²

iii. (12a + 17b) × 4c = 12a × 4c + 17b × 4c
= 48ac + 68bc

iv. (4x + 5y) × (9x + 7y)
= 4x × (9x + 7y) + 5y × (9x + 7y)
= (4x × 9x) + (4x × 7y) + (5y × 9x) + (5y × 7y)
= 36x² + 28xy + 45xy + 35y²
= 36x² + 73xy + 35y²

Question 2.
A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area. Solution:
Length of the rectangle = (8x + 5) cm
Breadth of the rectangle = (5x + 3) cm
∴ Area of the rectangle = length × breadth
= (8x + 5) × (5x + 3)
= 8x × (5x + 3) + 5 × (5x + 3)
= (8x × 5x) + (8x × 3) + (5 × 5x) + (5 × 3)
= 40x² + 24x + 25x + 15
= 40x² + 49x + 15
∴ The area of the rectangle is (40x² + 49x + 15) sq. cm.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 34 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Algebraic Expressions and Operations on them Class 7 Practice Set 34 Answers Solutions Chapter 8

Question 1.
Subtract the second expression from the first.
i. (4xy – 9z); (3xy – 16z)
ii. (5x + 4y + 7z); (x + 2y + 3z)
iii. (14x² + 8xy + 3y²); (26x² – 8xy – 17y²)
iv. (6x² + 7xy + 16y²); (16x² – 17xy)
v. (4x + 16z); (19y – 14z + 16x)
Solution:
i. (4xy – 9z) – (3xy – 16z)
= 4xy – 9z – 3xy + 16z
= (4xy – 3xy) + (16z – 9z)
= xy + 7z

ii. (5x + 4y + 7z) – (x + 2y + 3z)
= 5x + 4y + 7z – x – 2y – 3z
= (5x – x) + (4y – 2y) + (7z – 3z)
= 4x + 2y + 4z

iii. (14x² + 8xy + 3y²) – (26x² – 8xy – 17y²)
= 14x² + 8xy + 3y² – 26x² + 8xy + 17y²
= (14x² – 26x²) + (8xy + 8xy) + (3y² + 17y²)
= -12x² + 16xy + 20y²

iv. (6x² + 7xy + 16y²) – (16x² – 17xy)
= 6x² + 7xy + 16y² – 16x² + 17xy
= (6x² – 16x²) + (7xy + 17xy) + 16y²
= -10x² + 24xy + 16y²

v. (4x + 16z) – (19y— 14z + 16x)
= 4x + 16z – 19y + 14z – 16x
= (4x – 16x) – 19y + (16z + 14z)
= -12x – 19y + 30z

Balbharti Maharashtra State Board Class 7 English Solutions Chapter 2.7 Great Scientists Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 7 English Solutions Chapter 2.7 Great Scientists

Class 7 English Chapter 2.7 Great Scientists Textbook Questions and Answers

1. Write what is implied in the following sentences.

Question a.
But few know his inspirational life story, which is all about courage and fighting against the odds.
(What does it tell you about Faraday’s life?)
Answer:
Faraday was bom into an extremely poor family. He had a speech defect which ended his formal education. He wanted to become a scientist. He came across a lot of difficulties and failures but through courage and determination, he achieved fame and name.

Question b.
Even then Davy did not have much hope for Faraday.
(What do the words ‘even then’ suggest?)
Answer:
Davy hired Faraday as his Secretary. Faraday worked relentlessly and became indespensible to Davy. ‘Even then’ suggests that even though Faraday did everything to prove himself, Davy did not believe that Faraday was capable of doing anything great.

Question c.
People started telling Davy that of all his discoveries, the best was Faraday himself.
(What does it suggest about Davy’s work?)
Answer:
The line suggests that the people felt that Davy had made discoveries which were good but Faraday’s discoveries stood out in comparison to Davy’s work.

2. Break the passage into convenient smaller sections. Give sub-headings or titles to each section.

Question 1.
Break the passage into convenient smaller sections. Give sub-headings or titles to each section.
Answer:

• 1st para: Michael Faraday, an inspiration
• 2nd para: Unhappy childhood
• 3rd para: Reading, an obsession
• 4th para: Twist in life
• 5th para: Magic of Electricity
• 6th para: Play of destiny
• 7th para: Fruits of labour
• 8th para: Revolution for mankind
• 9th para: A scientist is born
• 10th para: Challenges in life
• 11th para: Magic of magnets
• 12th para: A legend is born
• 13th para: Failure – a stepping stone to success
• 14th para: Obstacles – a challenge
• 15th para: Luck smiles again
• 16th para: Believe in yourself

3. List the different gadgets and instruments mentioned in the passage. Find more information about them (at least 3), using the Internet.

Question 1.
List the different gadgets and instruments mentioned in the passage. Find more information about them (at least 3), using the Internet.
Answer:
Gadgets and Instruments:
Fan, air conditioners, sewing machines, power tools, cars, trains, aeroplane engines, induction motors, telescopes, Bavarian glass, microscopes, electrical generators, light bulbs. Now you add to the list of gadgets and instruments. You can add your own too.

1. Microscope: It is an instrument used to see objects that are too small to be seen by the naked eye. Antonie Van Leeuwenhoek (1632-1723), a Dutch scientist, who in the late 17th century became the first man to make and use a real microscope. All microscopes came with a lens that can magnify 40 times the normal size.

2. Electrical generator: It was invented by Michael Faraday in 1831. It is a device that converts mechanical energy into electrical energy. Sources of mechanical energy include steam, turbines, gas turbines, water turbines, etc. It works on the principle of electromagnetic induction.

3. Fan: A fan is a machine used to create flow within a fluid, typically a gas such as air. The fan consists of a rotating arrangement of blades which act on the fluid. The fan was invented by Schuyler Skaats Wheeler in 1882.

4. Find out more about the following scientists with the help of the internet.
(a) Michael Faraday
(b) Humphry Davy
(c) Thomos Edison
(d) James Maxwell

Question 1.
Find out more about the following scientists with the help of the internet.
Answer:
(a) Michael Faraday: Michael Faraday (1791-1867) was an English (British) Scientist who contributed to the study of electromagnetism and electrochemistry. He was highly influenced by Humphry Davy and William Thomas Brande. He had notable awards to his credit, a few being Royal Medal, Copley Medal, Rumford Medal, Albert Medal.

(b) Humphry Davy: Humphry Davy (1778-1829) was a British scientist. It was he who found out that the inhalation of nitrous oxide produced surprising results, it came to be known as the laughing gas. He was awarded a medal by Napoleon and he identified the element iodine for the first time. He is remembered for his discoveries of sodium, potassium, calcium.

(c) Thomas Alva Edison: Thomas Alva Edison (1847-1931) an American inventor and businessman, has been described as America’s greatest inventor. He invented the photograph in 1877 because of which he came to be known as ‘The Wizard of Monto Park’. Thomas Edison had hearing problems during his childhood which persisted. The phonograph was the first machine that could record the sound of someone’s voice and play it back.

Edison recited the nursery rhyme ‘Mary had a little lamb’ and the phonograph played the words back to him. This was invented by a man whose hearing was poor and thought himself as deaf. It was Edison who designed a system of power plants. He has got 1093 patents to his credit. Many awards to his credit. Some of them are Franklin Medal, Albert Medal, Technical Grammy Award, etc.

(d) James Clerk Maxwell: James Clerk Maxwell (1831-1879) was a Scottish scientist in the field of mathematical physics His most notable work was to formulate the classical theory of electromagnetic radiation. He is known as the Father of Modern Physics. His other contributions included producing the first colour photograph taken in 1861 and many more. He earned an array of awards. To name a few, Rumford Medal, Keith and Hopkins prize, Adams prize, etc.

5. Language Study: Subject and predicate: A word or a phrase in a sentence that shows who or what does the action, is the subject. In the following sentences, the underlined part is the subject.
The baby is laughing.
The glass fell from the table.
The predicate Ls the part of a sentence that tells us something about the subject. The underlined parts of the following sentences show the predicates in them.
The baby is laughing.
The glass fell from the table.

Class 7 English Chapter 2.7 Great Scientists Additional Important Questions and Answers

Answer the following questions.

Question 1.
What is the meaning of:

Question i.
Reading became his obsession.
Answer:
He could think of nothing else but reading.

Question ii.
Electricity became a lifelong fascination.
Answer:
Electricity enchanted him throughout his life. Electricity was his passion throughout his life.

Question 2.
Why do you think Faraday’s friend gave him a free ticket to Davy’s programme?
Answer:
Faraday’s friend gave him a free ticket to Davy’s programme because he knew about Faraday’s fascination for electricity and also that he could not afford the ticket.

Question 3.
Give one example each to show that:

Question i.
Faraday was a good pupil.
Answer:
Davy tried to find out how an electric current could be applied continuously, but failed to figure it out. He asked Faraday to try his hand at it. Faraday set to work and within a few days, solved the problem.

Question ii.
Davy was not a good mentor.
Answer:
Davy was not happy at Faraday’s achievement but was jealous of Faraday.

Question 4.
An induction motor is a commonly used electrical machine. What examples of its uses are given here?
Answer:
The examples of electrical machine in which induction motor is commonly used are:

1. fans
2. air conditioning
3. sewing machines
4. photographs
5. power tools
6. cars
7. trains
8. aeroplane engines.

Question 5.
Why were Faraday’s drawings not accepted?
Answer:
Faraday’s drawing were without any mathematical equations as he did not know much about advanced mathematics and lacked formal education. Faraday’s drawing without mathematical equation could not be understood and so his drawings were not accepted.

Question 6.
Does it mean they were wrong?
Answer:
No, Faraday’s drawings were not wrong. It lacked mathematical equations because of which it could not be understood.

Reading Skills, Vocabulary and Grammar

Read the following passage and do the activities.

Simple Factual Questions:

Question 1.
State whether true or false.
i. Faraday’s Laws are about electricity.
ii. Faraday did not suffer speech defect as an adult.
Answer:
i. True
ii. True.

Complex Factual Questions:

Question 1.
Where was Michael Faraday bom?
Answer:
Michael Faraday was born into a poverty- stricken family in a dirty London suburb.

Question 2.
What came to an end when Michael Faraday was twelve?
Answer:
Michael Faraday’s formal education came to an end when he was twelve.

Question 3.
What were the odds against Faraday in his childhood?
Answer:
Faraday was born into an extremely poor family. He suffered from speech defect as a child. He could not even pronounce his own name. Other children laughed at him and even his teachers did not help him. when he was twelve, his mother was forced to pull him out from school. This put an end to his formal education.

Vocabulary:

Give the word meaning for.

Question 1.
‘mentor’
Answer:
A wise and trusted counsellor or teacher.

Question 2.
‘reverse engineer it’
Answer:
Taking apart an object and analyzing its components and workings in detail.

Word building.

Question i.
Give the noun form:
(a) inventor
(b) friend
(c) mysterious
(d) agree
Answer:
(a) invention
(b)friendship
(c) mystery
(d) agreement

Question ii.
Give the verb form:
(a) explosion
(b) comprehensive
(c) revolution
(d) success
(e) inspirational
(f) education
(g) modem
(h) courage
Answer:
(a) explode
(b) comprehend
(c) revolve
(d) succeed
(e) inspire
(f) educate
(g) modernize
(h) encourage

Question iii.
Give the adverb form:
(a) difficult
(b) secret
(c) random
(d) invisible
Answer:
(a) difficulty
(b) secretly
(c) randomly
(d) invisibly

Question iv.
Give the adjective form:
(a) communicate
(b) education
(c) experiment
(d) succeed
Answer:
(a) communicative
(b) educative
(c) experimental
(d) successful

Grammar:

Question 1.
Michael Faraday is regarded as one of the most distinguished scientists and inventors! (Add question tag)
Answer:
Michael Faraday is regarded as one of the most distinguished scientists and inventors isn’t it?

Personal Response:

Question 1.
Do you like reading inspirational stories?
Answer:
Yes, I do like reading inspirational stories. Reading inspirational stories motivates me to do something in life so that I too can gain fame and name while trying to make the world a better place to live in.

Read the following passage and do the activities.

Simple Factual Questions:

State whether the following statements are true or false.

Question i.
Faraday succeeded Davy as the head of the laboratory after Davy’s death.
Answer:
True

Question ii.
He was so engrossed in the lecture that he applaud with the rest of the crowd.
Answer:
False

Complex Factual Questions:

Question 1.
What was the subject of Davy’s lecture?
Answer:
The subject of Davy’s lecture was ‘The Mysterious force of Electric Fluid’.

Question 2.
What happens when an electric current is applied to a wire?
Answer:
When an electric current is applied to a wire, it causes that wire to behave like a magnet.

Question 3.
Why did Davy choose Faraday as his assistant?
Answer:
Davy was temporarily blinded by a chemical explosion which took place inside his lab. He needed an assistant with an excellent memory. He was reminded of Faraday and decided to make him his assistant.

Question 4.
Was Davy fair in his treatment of Faraday?
Answer:
No, Davy was unfair in his treatment of Faraday. He looked down upon Faraday because of his low social status and education. He told Faraday not to aspire for too.much and to stick to book-binding.

Question 5.
How did Faraday respond to that treatment?
Answer:
Davy’s treatment did not dishearten Faraday. Faraday was relentless. He worked day and night and learnt as much as he could and became indispensible to Davy.

Vocabulary:

Question 1.
Give the noun forms of the following.
i. comprehensive
ii. aspire
Answer:
i. comprehension
ii. aspiration.

Question 2.
Give the adverb form:
i. difficult
ii. invisible
Answer:
i. difficulty
ii. invisibly.

Grammar:

Question 1.
Destiny had a strange plan in store for him. (Use a question tag)
Answer:
Destiny had a strange plan in store for him, didn’t it?

Question 2.
Identify the subject and predicate from the following sentences.

1. Davy became his role model.
2. Destiny had a strange plan in store for him. Subject Predicate

Subject	Predicate
a. Davy	became his role model.
b. Destiny	had a strange plan in store for him.

Personal Response:

Question 1.
What actions and thoughts of Faraday show that he was inspired by Davy?
Answer:
The actions and thoughts of Faraday which shows he was inspired by Davy are as follows:

1. He kept taking notes.
2. He was so engrossed in the lecture that he forgot to applaud.
3. He bound the notes he had taken down into a book, hoping to gift it to Davy.
4. He was inspired to become a great scientist and Davy became his role model.
5. Faraday thought that it would be wonderful if Davy became his mentor.

Read the following passage and do the activities.

Simple Factual Questions:

Say whether the following statements are true or false.

Question i.
Faraday succeeded Davy in 1829.
Answer:
True

Question ii.
In 1830 Faraday developed memory loss which continued for the rest of his life.
Answer:
False

Complex Factual Questions:

Question 1.
When did Faraday develop memory loss?
Answer:
Faraday developed memory loss in 1840.

Vocabulary:

Question 1.
Give a sub-heading or title to the extract.
Answer:
‘Faraday – a Legend is born’.

Grammar:

Question 1.
Find any 4 verbs from the above extract.
Answer:
died, succeeded, made, moved.

Question 2.
Find any 2 proper nouns from the above extract.
Answer:
Faraday and Davy.

Question 3.
Faraday was now a legend. (Add a question tag)
Answer:
Faraday was now a legend, wasn’t he?

Question 4.
Remember that piece of Bavarian glass Faraday had kept on his shelf? (State the Kind of sentence)
Answer:
Interrogative Sentence.

Personal Response:

Question 1.
Write the idea behind an electrical generator.
Answer:
Faraday noticed that if he moved a magnet, it would produce electrical current, converting motion into electricity. This is how the electrical generator came into being.

Question 2.
How far did Faraday convert the reminder of his first major failure into an instrument of great success?
Answer:
Faraday used the reminder of his first major failure, the Bavarian glass, to show the concept of polarisation. He thus converted it into an instrument of great success.

Question 3.
What complicated experiment did Faraday begin after he developed a memory loss?
Answer:
After developing a memory loss, Faraday began a complicated experiment to prove that light was closely related to electricity and magnetism.

Language Study

Do as directed.

Question 1.
Pick out the subject and predicate.

Question i.
A wealthy, educated physicist, he met.
Answer:
Subject: He
Predicate: met a wealthy, educated physicist

Question ii.
He was unable to explain.
Answer:
Subject: He
Predicate: was unable to explain

Question iii.
The induction motor spurred a revolution.
Answer:
Subject: The induction motor
Predicate: spurred a revolution

Question iv.
At thirteen, however, he stated working.
Answer:
Subject: He
Predicate: started working at thirteen

Question v.
One day he came across a book on electricity.
Answer:
Subject: He
Predicate: came across a book on electricity one day.

Question vi.
Other children laughed at him.
Answer:

• Subject: Other children
• Predicate: laughed at him
• Remember: If there is no subject, put a ‘you’ in the subjects place.
• e.g. Shut the door.
• Subject: You Predicate: Shut the door

Question 2.
Other children laughed at him. (Add a question tag)
Answer:
Other children laughed at him. didn’t they?

Question 3.
He started reading it. (State the kind of verb – Transitive/ Intransitive)
Answer:
started – transitive verb reading – transitive verb

Question 4.
Other children laughed at him. (Begin with’He…’)
Answer:
He was laughed at by other children.

Question 5.
Faraday solved the problem. (End with ‘…Faraday’)
Answer:
The problem was solved by Faraday.

Question 6.
Faraday was now a legend. (Make it interrogative)
Answer:
Wasn’t Faraday now a legend?

Question 7.
James Maxwell was a wealthy, educated physicist. (Frame a ‘Wh’ question so as to get the underlined word as the answer)
Answer:
Who was James Maxwell?

Question 8.
Get me a glass of water. (Separate the subject and predicate)
Answer:
Subject-You
Predicate – Get me a glass of water.

Question 9.
He undertook it as challenges and opportunities. (Identify the part of speech of the underlined words)
Answer:
challenges, opportunities – nouns.
Writing Skills

Question 10.
Find out from the internet an inspiring story.
Answer:
Lionel Messi, bom in 1987, into a middle-class family who lived in Argentina. He grew up in a football-loving family. He developed a passion for football at an early age. At 5, he showed great skills at a club coached by his father. It was strange to see that he was way too short when compared to the boys of his age. Wherever he went he was the shortest of all.

This did not stop him from playing. At the age of 11, he was diagnosed with growth hormone deficiency and had to take medical assistance. Every night he had to take an injection in his legs for 3 years. His family could not afford the treatment for long. Though he was a great asset for his soccer club, they refused to pay his bills. Lady Luck smiled on him and his family. At 13, he got an offer from F. C. Barcelona that they would pay for his treatment if he played for them.

His family moved to Spain with no money or job, all in the hope that his treatment will make him big enough to play soccer. At 17, despite being injury prone he became the youngest player in F. C. Barcelona to score a goal and became a star performer. Today, Lionel Messi is known as one of the greatest footballers of all time. He has won the most Ballon d’Ors in the world. The FIFA Ballon d’Or (Golden Ball) also known as the European Footballer of the year, was an annual association football award presented to the world’s best male player.
Messi’s story teaches us ‘Never lose hope.
You never know what tomorrow might bring. Keep believing in yourself’.

Great Scientists Summary in English

The lesson ‘Great Scientists’ is an inspirational and motivational lesson. The story about Michael Faraday and his journey through all odds in life to become one of the most distinguished scientists and inventors of modern times is really an eye-opener to all those who give up easily when faced with a difficult situation. Perseverance and a will to win will surely reach you to the height of success is described in this story. This is a story all teens must-read.

Introduction:

The lesson ‘Great Scientists’ is taken from the introduction to ‘Reignited’ by A. P. J. Abdul Kalam and Srijan Pal Singh.

Glossary:

1. persevering (v) – to continue doing something in a determined way despite having difficulties ‘
2. deterred (v) – discouraged
3. distinguished (adj) – renowned, famous
4. inspirational (adj) – encouraged or making you feel you want to do something
5. poverty-stricken (adj) – extremely poor
6. defect (n) – lack or fault in a person
7. formal education (n) – classroom based education given by trained teachers
8. obsession (n) – the state of being preoccupied with someone or something
9. fascination (n) – passion
10. renowned (adj) – famous
11. comprehensive (adj) – exhaustive
12. dejected (adj) – sad
13. relentless (adj) – persistent
14. indispensable (adj) – that one cannot do without
15. career (n) – profession, occupation
16. revolution (n) – something that signifies a great change
17. reverse (v) – to turn something inside out
18. accomplish (v) – to finish successfully
19. significant (n) – notable
20. souvenir (n) – an item of sentimental value
21. complicated (adj) – difficult
22. randomly (adj) – without definite direction
23. filings (n) – particles of something (metal)
24. invisible (adj) – which cannot be seen
25. galaxies (n) – milky way
26. aliens (n) – any life form of extra terrestrial origin
27. conviction (n) – belief
28. epitomizes (v) – sums up
29. mentor (n) – a wise and trusted counsellor or teacher
30. reverse engineer it – taking apart an object and analyzing its components and working in detail
31. dynamo – a machine for converting mechanical energy into electrical energy
32. applaud – clap.
33. perseverance – hard work.
34. aspirations (n) – hopes of achieving something.
35. binding (v) – to join separate pieces of paper and make a book.
36. chemist (n) – a person who does chemical experiments or research.
37. consistent (adj) – that does not change over a period of time.
38. engrossed (v) – engaged with complete attention.
39. hooked (adj) – being so excited about something that you are unable to leave.
40. isolated (adj) – alone, separate.
41. legend (n) – a person who is very famous and admired.
42. novel (adj) – new and unusual.
43. pursue (v) – to engage in an activity over a long period of time.

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Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 7 Motion, Force and Work Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 7 Science Solutions Chapter 7 Motion, Force and Work

Class 7 Science Chapter 7 Motion, Force and Work Textbook Questions and Answers

1. Fill ¡n the blanks with the proper words from the brackets.
(stationary, zero, changing, constant, displacement, velocity, speed. acceleration, stationary but not zero. inc reuses)

Question a.
If a body traverses a distance in direct proportion to the time, the speed of the body is ……………… .
Answer:
constant

Question b.
If a body is moving with a constant velocity, its acceleration is ……………… .
Answer:
zero

Question c.
……………. is a scalar quantity.
Answer:
Speed

Question d.
…………….. is the distance traversed by a body in a particular direction in unit time.
Answer:
Velocity

2. Observe the figure and answer the questions.

Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?

Question a.
Observe the figure and answer the questions

Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?
Answer:
1. Actual distance = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\) + \(\overline{\mathrm{CD}}\) + \(\overline{\mathrm{DE}}\) = 3 + 4 + 5 + 3
Actual distance = 15 km

2. Displacement = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BD}}\) + \(\overline{\mathrm{DE}}\)
= 3 + 3 + 3
Displacement = 9 km

3. Speed = \(\frac{\text { Distance travelled }}{\text { Total time }}\)
Distance = 15 km = 15 × 1000 = 15000 m
Time = 1 hr = 1 × 60 × 60 = 3600 sec.
s = \(\frac{15000}{3600}\) or s = \(\frac{15 \mathrm{~km}}{1 \text { hour }}\) = 15km/hour
= 4.16 m/sec. or 15 km/hour

4. Velocity = \(\frac{\text { Distance travelled }}{\text { Total time }}\)
Displacement = 9 km = 9 × 1000 = 9000 m
Time = 1 hr = 1 × 60 × 60 = 3600 sec
V = \(\frac{9000}{3600}\) or V = \(\frac{9 \mathrm{~km}}{1 \text { hour }}\) = 9 km/hour
= 2.5 m/sec. or 9 km/hour

5. Yes, this velocity can be called as average velocity.

3. From the groups B and C, choose the proper words, for each of the words in group A.

Question a.
From the groups B and C, choose the proper words, for each of the words in group A.

Answer:

Group ‘A’	Group B’	Group ‘C’
Work	Joule	erg
Force	Newton	dyne
Displacement	Metre	cm

4. A bird sitting on a wire, flies, circles around and comes back to its perch. Explain the total distance it traversed during its flight and its eventual displacement.

Question a.
Answer:
The total distance the bird has traversed is the length of the distance covered by circling, but the eventual displacement are the bird is zero as its initial and final position are one and the same.

5. Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.

Question a.
Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.
Answer:
1. Force: The interaction that brings about the acceleration is called force.
e.g: An ox is pulling a cart, applying brakes to a bicycle, lifting heavy iron object with a crane.

2. Work: When an object is displaced by applying a force on it, work is said to be done.
e.g: A bucketful of water is to be drawn from a well and taken to the home by walking from well to home.

3. Displacement: The minimum distance
traversed by a moving body in one direction from the original point to reach the final point is called displacement.
e.g: A rolling of a ball from point A to point B in the same direction.

4. Velocity: Velocity is the distance traversed by a body in a specific direction in unit time.
e.g: A truck is covering a distance of 40km from A to D in a straight line in 1 hour.

5. Acceleration: It is change in velocity per second. It can be deduced.
Acceleration = \(\frac{\text { Change in velocity }}{\text { Time taken for change }}\)
e.g:

(i) In the above example a truck covered the distance AB at velocity of 60 km/hr, BC at 30 km/hr and CD at 40 km/hr. (ii) It means that the velocity for the distance CD is greater than the velocity for the distance BC. (iii) From the number of seconds required for this change in velocity to take place, the change in velocity per second can be deduced. This is called acceleration (iv) Distance: The length of the route actually traversed by a moving body irrespective of the direction is called distance.
e.g: Ranjit travelled 1km. from his home to school.

6. A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.

Question a.
A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.

Answer:
As its initial and final positions are one and the same.
Initial Velocity = 2 cm/s.
Final Velocity = 4 cm/s
Time taken for the change in velocity from B to
D = 4 cm/s – 2 cm/s = 2 cm/s

7. Solve the following problems.

Question a.
A force of 1000 N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 10m. How much is the work done?
Answer:
Force (F) = 1000 N
displacement (s) = 10m
work done (W) = ?
W = Fs
= 1000 × 10
W = 10,000 Joule

Question b.
A cart with mass 20 kg went 50 m in a straight line on a plain and smooth road when a force of 2 N was applied to it. How much work was done by the force?
Answer:
Force (F) = 2 N
Displacement (s) = 50 m
Work done (W) = ?
W = Fs
= 2 × 50
W = 100 Joule

Project:

Question a.
Collect information about the study made by Sir Isaac Newton regarding force and acceleration and discuss it with your teacher.

Class 7 Science Chapter 7 Motion, Force and Work Important Questions and Answers

Fill in blanks:

Question 1.
Displacement is a …………. quantity.
Answer:
vector

Question 2.
The …………. of an object can change even while it is moving along a straight line.
Answer:
velocity

Question 3.
The …………. velocity can be different at different times.
Answer:
instantaneous

Question 4.
Change in velocity per second is called …………. .
Answer:
acceleration

Question 5.
The interaction that brings about the acceleration is called …………. .
Answer:
force

Question 6.
The scientist …………. was the first to study force and the resulting acceleration.
Answer:
Sir Isaac Nezvton

Question 7.
Ability to do work is called …………. .
Answer:
Energy

Question 8.
W = …………. × S.
Answer:
F

Question 9.
Unit of work is …………. and …………. .
Answer:
Joule, erg

Question 10.
Unit of force is …………. and …………. .
Answer:
Newton, dyne

Question 11.
Force is a …………. quantity.
Answer:
vector

Question 12.
The velocity at a particular time is called …………. velocity.
Answer:
instantaneous

Question 13.
The …………. of a body is the distance traversed per unit time.
Answer:
speed

Question 14.
Unit of acceleration is …………. and …………. .
Answer:
m/s² and cm/s²

Question 15.
Force is measured by the …………. that it produces.
Answer:
acceleration

Question 16.
Work done by a body with no displacement will be …………. .
Answer:
zero

Say whether True or False, correct the false 1 statements:

Question 1.
Velocity is distance travelled per unit of time.
Answer:
False. Speed is distance travelled per unit of time

Question 2.
In displacement, both distance and direction are taken into account.
Answer:
True

Question 3.
Speed = Distance/time.
Answer:
True

Question 4.
Change in speed per second is acceleration.
Answer:
False. Change in velocity per second is acceleration

Question 5.
Work done depends on the force and the displacement.
Answer:
True

Question 6.
C.G.S. unit of acceleration is m/s².
Answer:
False. C.G.S. unit of acceleration is cm/s².

Question 7.
M.K.S. unit of force is dyne.
Answer:
False. M.K.S. unit of force is Newton

Question 8.
Force is measured by the acceleration that it produces.
Answer:
True

Write the difference between the following:

Question 1.
Speed and Velocity
Answer:

Speed	Velocity
1. Speed is distance travelled per unit of time.	1. Velocity is the distance traversed by a body in a specific direction in unit time.
2. It is a scalar quantity.	2. It is a vector quantity.
3. Formula: Speed = \(\frac{\text { Distance traversed }}{\text { Total time }}\)	3. Formula: Velocity = \(\frac{\text { Displacement }}{\text { Total time }}\)

Question 2.
Distance and Displacement
Answer:

Distance	Displacement
1. The length of the route actually traversed by a moving body, irrespective of the direction is called distance.	1. The minimum distance traversed by a moving body in one direction from the original point to reach the final point is called displacement.
2. It is a scalar quantity.	2. It is a vector quantity.

Solve the following problems!

Question 1.
A bus travelled 200 km in the first 3 hours and then 100 kms for the next one and a half hours and then 120 kms for the next one and a half hours. What is the average velocity of the bus if it has moved in a straight line for the whole journey.
Answer:

Question 2.
See the diagram and calculate the Distance and Displacement travelled by the body from A to I.

Answer:
Distance travelled =
A → B → C → D → E → F → G → H + I
= 5 + 7 + 6 + 3 + 5 + 4 + 6 + 5
= 41 m
Displacement = A → I in a straight line shortest distance
= 1m

Use your brainpower:

Question 1.
The unit of acceleration is m/s², verify this.
Answer:

Question 2.
Acceleration is a vector quantity. Is force a vector quantity too?
Answer:
Yes, acceleration and force both are vector quantities, because both can be expressed completely only when magnitude and direction are given and the quantity which needs direction and magnitude both is called a vector quantity.