Author name: Prasanna

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1.
A plastic box 13 m long, 1.25 wide and 65 cm deep are to be made. It is to be open at the top. Ignoring the thickness of a plastic sheet, determine.
(i) The area of the sheet required for making the box.
(ii) The cost of the sheet for it, if a sheet measuring 1m costs Rs. 20.
Solution:
(i) We have given,
length of box = 1.5 m
width of the box = 1.25 m
height of the box = 65 cm = 0.65 m

∴ Surface area of box = 2(lb + bh + lh)
= 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5)
= 2(1.875 + 0.8125 + 0.975)
= 2 × 3.6625
= 7.325 m²
But we have given box is open at the top.
Area of top of the box = 1.875 m²
∴ Surface area of required box = 7.325 – 1.875 = 5.45 m²
So, the area of sheet required for making the box is 5.45 m²
(ii) Rate of sheet = Rs. 20/cm²
∴ Total cost of sheet for box = 20 × 5.45 = Rs. 109.

Question 2.
The length, breadth, and height of a room are 5 m, 3 m, and 3 m respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs. 7.50 per m².
Solution:

We have given
length of room = 5 m
breadth of a room = 4 m
height of a room = 3 m
Surface area of 4 walls and ceiling of the room
= 2(l + b)h + lb
= 2(5 + 4)3 + 5 × 4
= 54 + 20
= 74 m²
Cost of whitewashing the walls and ceiling of the room at Rs. 7.50 per m² is = 74 × 7.50 = Rs. 555.

Question 3.
The floor of the rectangular hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m² is Rs. 15000, find the height of the hall.
Solution:
We have given that
Perimeter of floor = 250 m
2(l + b) = 250
l + b = 125 m
Now,
Area of four walls = 2 (l + b)h
= 2 × 125 × h
= 250h m².
Rate of painting the four walls is Rs. 10 per m² is Rs. 250h × 10
But we have given that the total cost of painting of four walls = 15000
i.e. 2500h = 15000
or, h = \(\frac{15000}{2500}\) = 6 m
∴ Height of the hall = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimension 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution:
We have given that the container contains sufficient paint for the area of 9.375 m².

Again
length of the bricks = 22.5 cm = 0.225 m
breadth of the bricks = 10 cm = 0.10 m
height of the bricks = 7.5 cm = 0.075 m
∴ surface area of one brick = 2(lb + bh + lh)
= 2(0.225 × 0.10 + 0.10 × 0.075 + 0.075 × 0.225)
= 2(0.0225 + 0.0075 + 0.016875)
= 0.09375 m².
Area of one brick = 0.09375 m².
Let x bricks can be painted out of this container
∴ 0.9375 × x = 9.375
or, x = \(\frac{9.375}{0.09375}\) = 100
Total number of bricks can be painted out of this container is 100.

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has a greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
(i) Length of edge of cubical box = 10 cm
The lateral surface area of cubical box = 4a²
= 4 × (10)²
= 400 cm²

Again, length of cuboidal box = 12.5 cm
breadth of cuboidal box = 10 cm
and height of cuboidal box = 8 cm.
∴ Lateral surface area of cubodical box = 2(l + b)h
= 2(12.5 + 10)8
= 360 cm²
Lateral surface area of cubical box is greater than cubodical box by = (400 – 360) = 40 cm²
Now,
Total surface ara of cubical box = 6a²
= 6 × (10)²
= 600 cm²
and total surface area of cubodical box = 2 (lb + bh + lh)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5)
= 2(125 + 80 + 100)
= 610 cm²
Total surface area of cubodical box is greater than cubical box by = (610 – 600) = 10 cm².

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of glass?
(ii) How much tape is needed for all the 12 edges?
Solution:
(i) We have given that
length of small indoor greenhouse = 30 cm.
the breadth of small indoor greenhouse = 25 cm.
and height of small indoor greenhouse = 25 cm.

Total surface area of indoor greenhouse (herbarium)
= 2(lb + bh + lh)
= 2(30 × 25 + 25 × 25 + 25 × 30)
= 2(750 + 625 + 750)
= 4250 cm²
∴ Area of glass to make herbarium is 4250 cm².
(ii) Total length of tape is needed = 4(l + b + h)
= 4(30 + 25 + 25)
= 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs. 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Dimensions of the bigger box are 25 cm × 20 cm × 5 cm
Total surface ara of one big box = 2(lb + bh + lh)
= 2(25 × 20 + 20 × 5 + 5 × 25)
= 2(500 + 100 + 125)
= 1450 cm².
∴ Total surface area of 250 such boxes are 250 × 1450 = 3,62,600 cm².
Again,
Dimension of smaller box is 15 cm × 12 cm × 5 cm.
∴ Total surface area of one small box = 2(lb + bh + lh)
= 2(15 × 12 + 12 × 5 + 5 × 15)
= 2(180 + 60 + 75)
= 630 cm².
∴ Total surface area of 250 such boxes = 250 × 630 = 1,57,500 cm².
∴ Total surface area of both type of boxes are = 3,62,500 + 1,57,500 = 5,20,000 cm²
Now, 5% of the total surface area is required for the overlaps.
5% of 5,20,000 = \(\frac{5}{100}\) × 5,20,000 = 26,000
∴ Total area of card board required = 5,20,000 + 26000 = 5,46,000
∴ Total cost of cardboard at the rate of Rs. 4 per 1000 cm² = \(\frac{546000}{1000} \times 4\) = Rs. 2184

Question 8.
Praveen wants to make a temporary shelter for her car, by making a box-like structure with a tarpaulin that covers all four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5m, with base dimensions 4 m × 3 m?
Solution:
We have given that,
length of temporary shelter for car = 4 m
the breadth of temporary shelter for car = 3 m
and height of temporary shelter for car = 2.5 m

∴ Total tarpaulin be required to make the shelter = Area of 4 wall + Area of roof
= 2(l + b)h + l × b
= 2(4 + 3) × 2.5 + 4 × 3
= 2 × 7 × 2.5 + 4 × 3
= 35 + 12
= 47 m²
Therefore total tarpaulin be required to make temporary shelter for the car is 47 m².

These NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.1

Question 1.
Following number line shows the temperature in degree Celsius (°C) at different places on a particular day.

(a) Observe this number line and write the temperature of the places marked on it.
(b) What is the temperature difference between the hottest and the coldest places among the above?
(c) What is the temperature difference between Lahulspiti and Srinagar?
(d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?
Answer:

(b) Temperature difference between the hottest and the coldest places
= Temperature of Bengaluru – Temperature of Lahulspiti
= 22°C – (- 8°C) = 22°C + 8°C
= 30°C

(c) Temperature difference between Lahulspiti and Srinagar
= Temperature of Srinagar – Temperature of Lahulspiti
= – 2°C – (- 8°C) = – 2°C + 8°C = 6°C

(d) Temperature of Srinagar and Shimla together
= Temperature of Srinagar + Temperature of Shimla
= – 2°C + 5°C = 3°C
3°C < 5°C
Yes, sum of temperatures at Srinagar and Shimla is less than the temperature at Shimla. Now, temperature of Srinagar = – 2°C
Also 3°C > – 2°C
No, sum of the temperatures at Srinagar and Shimla is not less than the temperature at Srinagar.

Question 2.
In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, – 5,- 10, 15 and 10, what was his total at the end?
Answer:
Score in 1st round = 25
Score in 2nd round = – 5
Score in 3rd round = – 10
Score in 4th round = 15
Score in 5th round = 10
Total score at the end
= 25 + (-5) + (-10) + 15 + 10
= 25 – 5 – 10 + 15 + 10
= 25 + 15 + 10 – 5 – 10
= 50 – 15 = 35

Question 3.
At Srinagar temperature was – 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?
Answer:
Temperature at Srinagar on Monday = -5°C
Drop in temperature at Srinagar on Tuesday = 2°C
∴ Temperature at Srinagar on Tuesday = – 5°C – 2°C = – 7°C
Rise in temperature at Srinagar on Wednesday = 4°C
∴ Temperature at Srinagar on Wednesday = – 7°C + 4°C = – 3°C

Question 4.
A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

Answer:
Since the sea level is at 0 m and the plane is 5000 m above the sea level. Also the submarine is 1200 m below the sea level, so vertical distance between the plane and submarine is
= 5000 m – (- 1200 m)
= 5000 m + 1200 m
= 6200 m

Question 5.
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹ 1, 642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.
Answer:
Amount deposited = ₹ 2000
Amount withdrawn = ₹ 1642
Balance in Mohan’s account = ₹ 2000 – ₹ 1642
after withdrawal = ₹ 358
Amount deposited is represented by positive integer.

Question 6.
Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?

Answer:
Distance towards west is represented by negative integer
The directions east and west are opposite to each other
Distance moved towards west = – 30 km
Distance moved towards east = + 20 km
Her final position from A = + 20 km + (- 30 km) = 20 km – 30 km
= – 10 km

Question 7.
In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square?
(i)

(ii)

Answer:
(i) Sum of the digits along
1st row = 5 + (- 1) + (- 4)
= 5 – 1 – 4 = 0
2nd row = – 5 + (- 2) + 7
= -5 – 2 + 7 = 0
3rd row = 0 + 3 + (- 3)
=0 + 3 – 3 = 0
1st column = 5 + (- 5) + 0
= 5 – 5 + 0 = 0
2nd column = -1 + (- 2) + 3
= -1 -2 + 3 = 0
3rd column = – 4 + 7 + (- 3)
= – 4 + 7- 3 = 7 – 7 = 0
1st diagonal = 5 + (- 2) + (- 3)
= 5 – 2 – 3 = 0
2nd diagonal = – 4 + (- 2) + 0
= – 4 – 2 + 0
= – 6 ≠ 0
∴ Square (i) is not a magic square.

(ii) Sum of the digits along
1st row = 1 + (- 10) + 0
= 1 – 10 + 0 = – 9
2nd row = – 4 + (- 3) + (- 2)
= – 4 – 3 – 2 = -9
3rd row = – 6 + 4 + (- 7)
= – 6 + 4 – 7 = – 13 + 4 = -9
1st column = 1 + (- 4) + (- 6)
= 1 – 4 – 6 = 1 – 10 = -9
2nd column = (- 10 ) + (- 3) + 4
= -10 -3 + 4 = – 13 + 4 = – 9
3rd column = 0 + (-2) + (-7)
= 0 – 2 – 7 = 0 – 9 = – 9
1st diagonal = 1 + (- 3) + (- 7)
= 1 – 3 -7 = 1 – 10 = – 9
2nd diagonal= – 6 + (- 3) + 0
= -9 + 0 = -9
∴ Sum of each row, column and diagonal is the same.
∴ The square (ii) is the magic square.

Question 8.
Verify a – (- b) = a + b for the following values of a and b.
(i) a = 21, b = 18
(ii) a = 118, b = 125 .
(iii) a = 75, b = 84
(iv) a = 28, b = 11
Answer:
(i) L.H.S = a – (-b) = 21 – (-18)
= 21 + 18 = 39
R.H.S = a + b = 21 + 18 = 39
∴ a – (- b) = a + b

(ii) L.H.S = a – (- b) = 118 – (-125)
= 118 + 125 = 243
R.H.S = a + b = 118 + 125 = 243
∴ a – (- b) = a + b

(iii) L.H.S = a – (- b) = 75 – (- 84)
= 75 + 84 = 159
R.H.S = a + b = 75 + 84 = 159
∴ a – (-b) = a + b

(iv) L.H.S = a – (- b) = 28 – (- 11)
= 28 + 11 = 39
R.H.S = a + b = 28 +11 = 39
∴ a – (- b) = a + b

Question 9.
Use the sign of >, < or = in the box make the statements true.


Answer:
(a) (- 8) + (- 4) = – 8 – 4 = -12
(- 8) – (- 4) = – 8 + 4 = – 4
∴ – 12 < – 4

(b) (-3) + 7 – (19) = -3 + 7 – 19
= – 22 + 7 = – 15
15 – 8 + (- 9) = 15 – 8 – 9
= 15 – 17 = – 2
∴ -15 < – 2

(c) 23 – 41 + 11 = 34 – 41 = – 7
23 – 41 – 11 = 23 – 52 = – 29
∴ – 7 > – 29

(d) 39 + (- 24) – (15) = 39 – 24 – 15
= 39 – 39 = 0
36 + (- 52) – (- 36) = 36 – 52 + 36
= 72 – 52 = 20
∴ 0 < 20

(e) – 231 + 79 + 51 = – 231 + 130 = – 101
– 399 + 159 + 81 = – 399 + 240 = – 159
∴ -101 > -159

Question 10.
A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.

(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?
(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?
(iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the 1 following:
(a) – 3 + 2 -… = – 8
(b) 4 – 2 + … = 8.
In (a) the sum (- 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
Answer:
(i) To reach the water level his jumps will T be as follows:
(- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) = – 8
Hence, in 11 jumps he will reach the 1 ninth step (water level)

(ii) To reach back the top step his jumps will be as follows:
4 + (-2) + 4 +(-2) + 4 = 8
Therefore, he will reach back the top step in 5 jumps.

(iii) (a) (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3 + 2) + (- 3) = – 8
(b)4 – 2 + 4 – 2 + 4 – 2 + 4 – 2 = 8
Also in (a) the sum (- 8) represent going down by 8 steps and in (b) the sum (+ 8) will represent going up by 8 steps.
OR
(i) The position of the monkey after the:
(a) 1st jump is at the fourth step.
(b) 2nd jump is at the 2nd step.
(c) 3rd jump is at the 5th step.
(d) 4th jump is at the 3rd step.
(e) 5th jump is at the 6th step.
(f) 6th jump is at the 4th step.
(g) 7th jump is at the 7th step.
(h) 8th jump is at the 5th step.
(i) 9th jump is at the 8th step.
(j) 10th jump is at the 6th step.
(k) 11th jump is at the 9th step.
∴ Required number of jumps = 11.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals InText Questions

NCERT In-text Question Page No. 165

Question 1.
Can you now write the following as decimals?

Answer:
We have,
(i) 5 hundreds + 3 tens + 8 ones + 1 tenth
= 5 x 100 + 3 x 10 + 8 x 1 + 1 x \(\frac { 1 }{ 10 }\)
= 500 x 30 + 8 + \(\frac { 1 }{ 10 }\)
= 538 x \(\frac { 1 }{ 10 }\) = 538.1

(ii) 2 hundreds + 7 tens + 3 ones + 4 tenths
= 2 x 100 + 7 x 10 + 3 x 1 + 4 x \(\frac { 1 }{ 10 }\)
= 200 x 70 + 3 + \(\frac { 4 }{ 10 }\)
= 273 x \(\frac { 4 }{ 10 }\) = 273.4

(iii) 3 hundreds + 3 tens + 4 ones + 1 tenth
= 3 x 100 + 5 x 10 + 4 x 1 + 6 x \(\frac { 1 }{ 10 }\)
= 300 x 50 + 4 + \(\frac { 6 }{ 10 }\)
= 354 x \(\frac { 6 }{ 10 }\) = 354.6

Question 2.
Write the lengths of Ravi’s and Raju’s pencils in ‘cm’ using decimals.
Answer:
10 mm = 1 cm
Therefore, 1 mm = \(\frac { 1 }{ 10 }\) cm
So, 5 mm = 5 x \(\frac { 1 }{ 10 }\) cm = \(\frac { 5 }{ 10 }\) cm
and 3 mm = 3 x \(\frac { 1 }{ 10 }\) cm = \(\frac { 3 }{ 10 }\) cm
Now, 7 cm 5 mm = 7 cm + 5 mm
= 7 cm + \(\frac { 5 }{ 10 }\) cm
= 7.5 cm
Again
8 cm 3 mm = 8 cm + 3 mm
= 8 cm + \(\frac { 3 }{ 10 }\)cm
= 8.3 cm

Question 3.
Make three more examples similar to the one given in question 1 and solve them.
Answer:
Please try yourself.

NCERT In-text Question Page No. 167

Question 1.
Write \(\frac{3}{2}, \frac{4}{5}, \frac{8}{5}\) in decimal notation.
Answer:
(i) \(\frac{3}{2}\)
We have, \(\frac{3}{2}=\frac{3 \times 5}{2 \times 5}=\frac{15}{10}=\frac{15}{10}\) = 1.5
∴ \(\frac{3}{2}\) = 1.5

(ii) \(\frac{4}{5}\)
We have, \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\) = 0.8
= \(\frac{8}{10}\) = 0.8
Thus, \(\frac{4}{5}\) = 0.8

(iii) \(\frac{8}{5}\)
We have, \(\frac{8}{5}=\frac{8 \times 5}{5 \times 5}=\frac{16}{10}\) = 1.6
Thus, \(\frac{8}{5}\) = 1.6

NCERT In-text Question Page No. 175

Question 1.
(i) Write 2 rupees 5 paise and 2 rupees 50 paise in decimals.
(ii) Write 20 rupees 7 paise and 21 rupees 75 paise in decimals.
Answer:
(i) (a) 2 rupees 5 paise :
2 rupees + 5 paise = 2 rupees + \(\frac{5}{10}\) rupees
∵ 100 paise = ₹1
∴ 1 paise = ₹ \(\frac{1}{100}\)
= (2 + 0.05) rupees = ₹2.05

(b) 2 rupees 50 paise =
2 rupees + 50 paise = 2 rupee + \(\frac{50}{100}\) rupees
∵ 100 paise = ₹1
∴ 1 paise = ₹ \(\frac{1}{100}\)
= (2 + 0.50) rupees = ₹ 2.50

(ii) (a) 20 rupees 7 paise :
20 rupees +7 paise
= 20 rupees + 7 x\(\frac{1}{100}\) rupees
[ ∴ \(\frac{1}{100}\) rupee = 1 paise]
= 20 rupees + 0.07 rupee
= ₹ (20 + 0.07)
= ₹ 20.07

(b) 21 rupees 75 paise :
21 rupees + 75 paise
= 21 rupees + 75 x -jy rupees
= 21 rupees + -jy rupees
= ₹ [21 + 0.75] = ₹ 21.75

NCERT In-text Question Page No. 176

Question 1.
Can you write 4 mm in ’em’ using decimals?
Answer:
Yes,
Since 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm
or 4 mm = 4 x \(\frac { 1 }{ 10 }\) cm = 0.4 cm

Question 2.
How will you write 7 cm 5 mm in ’em’ using decimals.
Answer:
7 cm 5 mm:
Since, 10 mm = 1 cm
∴ 1 mm= \(\frac { 1 }{ 10 }\) cm
Now, 7 cm 5 mm = 7 cm + 5 mm
= 7 cm + 5 x \(\frac { 1 }{ 10 }\) cm =[ 7 + \(\frac { 5 }{ 10 }\) cm
= (7 + 0.5) cm = 7.5 cm

Question 3.
How will you write 2008 m in ‘km’?
Answer:
Yes, we can change the given ‘metres’ into kilometres.
(a) 52 m
∵ 1000m = 1km ∴ 1m = \(\frac{1}{1000}\)
or 52 m = 52 x \(\frac{1}{1000}\) km = 0.052 km

(b) 340m
∵ 1m = \(\frac{1}{1000}\) km
∴ 340m = 340 x \(\frac{1}{1000}\) km
= \(\frac{340}{1000}\) km = 3.40 km

(c) 2008m
∵ 1 m = \(\frac{1}{1000}\) km
∴ 2008m = 2008 x \(\frac{1}{1000}\) km = \(\begin{aligned}
&2008 \\
&\hline 1000
\end{aligned}\) km

= (2 + 0.008) km = 2.008 km

Question 4.
Can you now write 52 m as ‘km’ using decimals? How will you write 340 m as ‘km’ using decimals?
Answer:
1000g = 1kg
∴ 1 g = \(\frac{1}{1000}\)kg
∴ 456g = \(\frac{1}{1000}\) x 456kg
= \(\frac{456}{1000}\) kg = 0.456kg

Question 5.
How will you write 2 kg 9 g in kg’ using decimals?
Answer:
2kg9g = 2kg + 9g
= 2kg + (9 x
= (2 + \(\frac{9}{1000}\))
= (2 + 0.009) kg = 2.009 kg

NCERT In-text Question Page No. 178

Question 1.
Find
(i) 0.29 + 0.36
(ii) 0.7 + 0.08
(iii) 1.54 + 1.80
(iv) 2.66 + 1.85
Answer:
(i) 0.29 + 0.36

∵ (9 + 6) hundredths =15 hundredths = 1
tenths + 5 hundredths
Thus, 0.29 + 0.36 = 0.65

(ii) 0.7 + 0.08

Thus, 0.7 + 0.08 = 0.78

(iii) 1.54 + 1.80

Thus, 1.54 + 1.80 = 3.34
Note: 5 tenths + 8 tenths =13 tenths
and 13 tenths = 10 tenths + 3 tenths = 1 one + 3 tenths

(iv) 2.66 + 1.85

Thus, 2.66 + 1.85 = 4.51

NCERT In-text Question Page No. 180

Question 1.
Subtract 1.85 from 5.46
Answer:
1.85 and 5.46 are ‘like decimals’.

Here, 1 is borrowed from ‘ones’ and given to tenths such that:
4 tenths + 10 tenths =14 tenths
5 ones – 1 one = 4 ones

Question 2.
Subtract 5.25 from 8.28
Answer:
5.25 and 8.28 are ‘like decimals’.

Question 3.
Subtract 0.95 from 2.29
Answer:
0.95 and 2.29 are ‘like decimals’.

Question 4.
Subtract 2.25 from 5.68
Answer:
2.25 and 5.68 are ‘like decimals’.

These NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2

Question 1.
A park in the shape of quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
In ABCD,
CD = 5 cm, BC = 12 and ∠C = 90°

∴ By pythagoras theorem we know that
h² = P² + b²
⇒ BD² = CD² + BC²
⇒ BD² = 5² + 12²
⇒ BD² = 169
⇒ BD = √169 = 13 cm.
Area of triangle BCD = \(\frac {1}{2}\) × 12 × 5 = 30 cm.
In ∆ABD,
AD = 8 m, AB = 9 m, and BD = 13 cm
∴ s = \(\frac{8+9+13}{2}\) = 15
∴ By Heron’s formula we know that
Area of triangle

= 35.5 m² (approx)
Now,
Area perpendicular ABCD = Ar ΔBCD + Ar ΔABD
= 30 + 35.5
= 65.5 m² (approx)
Therefore, quadrilateral ABCD occupy 65.5 m² area.

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
In ΔABC,
AB = 3 cm, BC = 4 cm and AC = 5 cm
∴ s = \(\frac{3+4+5}{2}\) = 6
∴ By Heron’s formula we know that
Area of ΔABC

Again in ΔACD,
AC = 5 cm, AD = 5 cm, and CD = 4 cm
∴ s = \(\frac{5+5+4}{2}\) = 7
∴ By Heron’s formula we know that,

= 9.2 cm² (approx)
∴ Area of quadrilateral ABCD = Ar ΔABC + Ar ΔACD
= 6 + 9.2
= 15.2 cm² (approx)

Question 3.
Radha made a picture of an aeroplane with coloured paper .is shown in Fig. 12.15. Find the total area of the paper used.

Solution:
In part I, sides of triangle are 5 cm, 5 cm and 1 cm
∴ s = \(\frac{5+5+1}{2}\) = 5.5
∴ By Heron’s formula

The area of the IInd part which is in the form of the rectangle is
6.5 × 1 = 6.5 cm² (Area of rectangle = l × b)
Area of IIIrd part = 3 × \(\frac{\sqrt{3}}{4}\) cm = 1.3 cm² (approx)
Area of IVth part, which is in the form of triangle is = \(\frac {1}{2}\) × 15 × 6 = 4.5 cm²
Area of Vth part, which is also in Hie form of triangle is = \(\frac {1}{2}\) × 15 × 6 = 4.5 cm²
Therefore, total area of the paper used = 2.5 + 6.5 +1.3 + 4.5 + 4.5 = 19.3 cm² (approx)

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm, and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
Sides of triangle are 26 cm, 28 cm and 30 cm.
∴ s = \(\frac{26+28+30}{2}\) = 42
∴ By Heron’s formula we know that

According to question,
Area of parallelogram = Area of triangle
Base × altitude = 336
28 × altitude = 336
∴ Altitude = \(\frac{336}{28}\) = 12
∴ Height of the parallelogram stands on the base 28 cm is 12 cm.

Question 5.
A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much rea of grass field will each cow be getting.

Solution:
In ∆ABD
AB = 30 m, AD = 30 m and BD = 48 m.
∴ s = \(\frac{30+30+48}{2}\) = 54
∴ By Heron’s formula we know that

∴ Area of rhombus ABCD = 2 × 432 = 864 m²
Therefore, area of grass getting by each cow = \(\frac{864}{18}\) = 48 m²

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:
For one triangular piece,
∴ s = \(\frac{20+50+50}{2}\) = 60
∴ By Heron’s formula
Area of one triangular piece

Therefore,
Area of five triangular piece = 5 × 200√6 = 1000√6 cm²
Hence, clothes of each colour are required for the umbrella is 1000√6 cm².

Question 7.
A kite in the shape of a square with a diagonal 32 cm and art isosceles triangle of base S cm and sides 6 cm each is lobe made of three diffident shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

Solution:
We have given that, the shape of a kite is square and length of its diagonal is 32 cm.
Area of 1 triangular region
= \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 32 × 16
= 256 cm²
(∵ We know that diagonals of square bisects each other at 90°)
Now, Area of IInd triangular region
= \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 32 × 16
= 256 cm²
Again, we have given that IIIrd part of the given kite is in the shape of an isosceles triangle whose equal sides are 6 cm and the base is 8 cm.
s = \(\frac{a+b+c}{2}\)
= \(\frac{6+6+8}{2}\)
= 10
∴ By Heron’s formula,
Area of IIIrd triangular region

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm, and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50 p per cm².

Solution:
The sides of one triangular tile are 9 cm, 28 cm, and 35 cm
s = \(\frac{9+28+35}{2}\) = 36 cm
By Heron’s formula, we know that
Area of one triangular tile

Therefore, area of 16 triangular tiles = 88.18 × 16 = 1410.88 cm²
Hence, the cost of polishing at the rate of 50 p per cm² is 1410.88 × 0.50 = Rs. 705.44 (approx)

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:
Draw a line through A and parallel to BC cut CD at E. Again draw AM ⊥ CD.
Now, sides of ∆ADE are 13 m, 15 m and 14 m
s = \(\frac{13+14+15}{2}=\frac{42}{2}\) = 21 m
By Heron’s formula, we know that

But, Area of ∆ADE = \(\frac {1}{2}\) × DE × AM
or, 84= \(\frac {1}{2}\) × 15 × AM
or, AM = \(\frac{84 \times 2}{15}\) = 11.2 m
We know that
Area of trapezium
= \(\frac {1}{2}\) × (25 + 10) × 11.2
= \(\frac {1}{2}\) × 35 × 11.2
= 196 m²

These NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1

Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using the Herons formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
In first case, length of each sides of an equilateral triangle is a
s = \(\frac{a+b+c}{2}=\frac{3 a}{2}\)
Therefore by Heron’s formula-

or, area of triangle = \(\frac{\sqrt{3} \cdot a^{2}}{4}\)
In case II,
Perimeter of an equilateral triange is 180 cm
∴ One side = \(\frac{180}{3}\) = 60 cm
s = \(\frac{60+60+60}{2}\) = 90 cm.
∴ Area of signal board = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{90(90-60)(90-60)(90-60)}\)
= \(\sqrt{90 \times 30 \times 30 \times 30}\)
= 900√3 cm²
or, Area of signal board = 900√3 cm².

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the wall are 122 m, 22 m, and 120 m (see fig. 12.9). The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:
The sides of the triangular wall are 122 m, 22 m, and 120 m.
∴ s = \(\frac{120+22+120}{2}\) = 132 m.
Therefore, by Heron’s Formula.
Area of triangular wall
= \(\sqrt{132(132-122)(132-122)(132-122)}\)
= \(\sqrt{132 \times 10 \times 110 \times 12}\)
= 1320 m²
∴ Rent = 5000 × 1320 × \(\frac{3}{12}\) = 16,50,000
Therefore, rent of this wall for 3 months is Rs. 16,50,000

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour.

Solution:
The sides of the wall are 15 m, 11 m, and 6 m.
∴ s = \(\frac{15+11+6}{2}\) = 16 m
∴ By Heron’s formula, we know that
Area of the wall

Therefore, the Area of the wall is 20√2 m².

Question 4.
Find the area of the triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Let third side of triangle is x cm.
∴ Perimeter = 18 + 10 + x
⇒ 42 = 18 + 10 + x
⇒ x = 14 cm
s = \(\frac{18+10+14}{2}=\frac{42}{2}\) = 21
By Heron’s formula we know that
Area of triangle

Therefore, the area of the triangle = 2√11 cm².

Question 5.
Sides of triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
Let the first side of triangle = 12x
Second side of triangle = 17x
and third side of triangle = 25x
Therefore,
perimeter of triangle = 12x + 17x + 25x
⇒ 540 = 54x
⇒ x = 10
First side of triangle = 12x = 12 × 10 = 120 cm
Second side of triangle = 17x = 17 × 10 = 170 cm
and Third side of triangle = 25x = 25 × 10 = 250 cm
∴ s = \(\frac{120+170+250}{2}=\frac{540}{2}\) = 270
∴ By Heron’s formula we know that
Area of triangle

Therefore, area of triangle = 9000 cm²

Question 6.
An isosceles triangle has a perimeter of 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Let the third side of an isosceles triangle is x.
∴ Perimeter of triangle = 12 + 12 + x
or, 30 = 24 + x
or, x = 6x m.
The third side of an isosceles triangle is 6 cm.
∴ s = \(\frac{12+12+6}{2}\) = 15
By Heron’s formula, we know that
Area of triangle

Therefore, the area of an isosceles triangle is 9√15 cm².

These NCERT Solutions for Class 8 Science Chapter 14 Chemical Effects of Electric Current Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Chemical Effects of Electric Current NCERT Solutions for Class 8 Science Chapter 14

Class 8 Science Chapter 14 Chemical Effects of Electric Current Textbook Exercise Questions and Answers

Page 180-181

Question 1.
Fill in the blanks:
a. Most liquids that conduct electricity are solutions of ……………., ……………. and …………….
b. The passage of an electric current through a solution causes ……………. effects.
c. If you pass current through copper sulphate solution, copper gets deposited on the plate connected to the ……………. terminal of the battery.
d. The process of depositing a layer of any desired metal on another material by means of electricity is called …………….
Answer:
a. acids, bases and salts
b. chemical
c. negative
d. electroplating

Question 2.
When the free ends of a tester are dipped into a solution, the magnetic needle shows deflection. Can you explain the reason?
Answer:
The deflection in the compass needle shows that current is flowing through the wound up wire and hence, through the circuit. This shows that the solution is a conducting solution because only then the current can flow through it through the circuit. This current through the wound up wire produces a magnetic field around it which acts on the magnetic needle of the compass and deflects it.

Question 3.
Name three liquids, which when tested in the manner shown in the figure, may cause the magnetic needle to deflect.

Answer:
Liquids like lemon juice, saltwater and vinegar can conduct electricity. Hence, these liquids can cause the magnetic needle to deflect.

Question 4.
The bulb does not glow in the setup shown in the figure. List the possible reasons. Explain your answer.

Answer:
The bulb may not glow because of the following reasons:

• Liquid in the beaker may be non-conducting. In such a case, the electric current would not be able to pass through the liquid. Hence, the circuit will not be complete.
• The conductivity of the liquid may be very low and so the current flowing through the circuit may be too weak to produce enough heat in the filament of the bulb to make it glow.
• The battery may be exhausted and may not have sufficient energy to generate electricity.
• The bulb may be fused or the circuit connections may be loose.

Question 5.
A tester is used to check the conduction of electricity through two liquids, labelled A and B. It is found that the bulb of the tester glows brightly for liquid A while it glows very dimly for liquid B. You would conclude that
a. liquid A is a better conductor than liquid B
b. liquid B is a better conductor than liquid A
c. both liquids are equally conducting
d. conducting properties of liquid cannot be compared in this manner
Answer:
a. liquid A is a better conductor than liquid B.

Question 6.
Does pure water conduct electricity? If not, what can we do to make it conducting?
Answer:
No, pure water does not conduct electricity. This is because pure water is devoid of any salts. Pure water can conduct electricity when a pinch of salt is added to it, as salt solution is conducting in nature.

Question 7.
In case of a fire, before the firemen use the water hoses, they shut off the main electrical supply for the area. Explain why they do this.
Answer:
Normal water is a good conductor of electricity. Therefore, firemen shut off the main electrical supply for the area before using the water hoses in the case of fire to prevent themselves from getting an electric shock.

Question 8.
A child staying in a coastal region tests the drinking water and also the seawater with his tester. He finds that the compass needle deflects more in the case of seawater. Can you explain the reason?
Answer:
Water we use for drinking purposes has very less amount of salts dissolved in it. However, seawater has very high concentration of salts in it and hence is a greater conductor of electricity than drinking water. Therefore, more current flows through seawater and the compass needle shows more deflection in this case.

Question 9.
Is it safe for the electrician to carry out electrical repairs outdoors during heavy downpour? Explain.
Answer:
Rainwater may contain many impurities which get dissolved in it when it passes through the various layers of troposphere. Due to this, rainwater becomes a good conductor of electricity. If an electrician carries out electrical repairs outdoors during heavy downpour, he may get an electric shock. Hence, it is not safe for the electrician to carry out electrical repairs outdoors during heavy downpour.

Question 10.
Paheli had heard that rainwater is as good as distilled water. So she collected some rainwater in a clean glass tumbler and tested it using a tester. To her surprise she found that the compass needle showed deflection. What could be the reasons?
Answer:
Rainwater is a pure water but many impurities and acidic gases get dissolved in the rainwater while it falls to the earth through the atmosphere. Due to the presence of these impurities and small amounts of acids in it, the rainwater conducts electricity. This explains Paheli’s observation that rainwater can allow electricity to pass through it while distilled water cannot.

Question 11.
Prepare a list of objects around you that are electroplated.
Answer:
Examples of electroplated objects are as follows:

• Chromium plating is done on different parts of cars, buses and motorcycles to give them shiny appearance.
• A fine layer of gold is deposited on the silver ornaments, and they are called gold-plated ornaments.
• Iron used in constructing a building is coated with a layer of zinc (galvanisation). This protects iron from corrosion and rusting.

Question 12.
The process that you saw in Activity 14.7 (of NCERT textbook) is used for purification of copper. A thin plate of pure copper and a thick rod of impure copper are used as electrodes. Copper from impure rod is sought to be transferred to the thin copper plate. Which electrode should be attached to the positive terminal of the battery and why?
Answer:
The thin plate of pure copper should be connected to the positive terminal of the battery. This is because when electric current is passed through the copper sulphate solution, it dissociates into copper and sulphate. The free copper is drawn towards the negative terminal of the battery and gets deposited on it. On the other hand, the loss of copper from the solution would be regained from the impure copper rod which is connected to the positive terminal of the battery.

NCERT Extended Learning Activities and Projects

Question 1.
Test the conduction of electricity through various fruits and vegetables. Display your result in a tabular form.
Hint:
Fruits such as oranges, apples, peach and grapes are good conductors of electricity whereas fruits like kiwi, banana, papaya, pineapple are poor conductors of electricity. On the other hand, vegetables such as lemon, tomatoes, carrot and reddish are good conductors of electricity whereas onion, cabbage, cauliflower are poor conductors of electricity.

Question 2.
Repeat Activity 14.7 (of NCERT textbook) with a zinc plate in place of the copper plate connected to the negative terminal of the battery. Now replace zinc plate with some other metallic object and again repeat the activity. Which metal gets deposited over which other metal? Discuss your findings with your friends.
Hint:
In each case copper gets deposited over the metal plate because the electrolyte used is copper sulphate and the positive electrode is of copper.

Question 3.
Find out if there is a commercial electroplating unit in your town. What objects are electroplated there and for what purpose? (The process of electroplating in a commercial unit is much more complex than what we did in Activity 14.7). Find out how they dispose off the chemicals they discard.
Hint:
Do it yourself.

Question 4.
Imagine that you are an ‘entrepreneur’ and have been provided a loan by a bank to set up a small electroplating unit. What object would you like to electroplate and for what purpose? (Look up the meaning of ‘entrepreneur’ in a dictionary).
Hint:
If I were an entrepreneur and was provided a loan by a bank to set up a small electroplating unit, I would start an electroplating unit of electronic and electrical appliances because:

• There is an increasing demand for electronic devices.
• The telecommunication industry is ever-expanding.
• Cost efficiency of electroplating these articles is high as compared to other technologies.

Question 5.
Find out the health concerns associated with chromium electroplating. How are people trying to resolve them?
Hint:
The health problems associated with chromium electroplating are skin rashes, stomach ulcers, respiratory problems, weak immunity, kidney and liver damage and cancer. Steps taken to resolve it may include use of water insoluble chromium compounds and mild steel.

Question 6.
You can make a fun pen for yourself. Take a conducting metal plate and spread a moist paste of potassium iodide and starch. Connect the plate to a battery as shown in the figure. Now using the free end of the wire, write a few letters on the paste. What do you see?

Hint:
Do it yourself.

Activity 1

Objective: To test the electrical conductivity of lemon juice (acid) or vinegar.
Materials Required: A few small plastic or rubber caps of discarded bottles, lemon juice/ vinegar, tester.
Procedure:

• Collect a few small plastic or rubber caps of discarded bottles and clean them.
• Pour one teaspoonful of lemon juice or vinegar in one cap.
• Bring your tester over this cap and let the ends of the tester dip into the lemon juice or vinegar. Take care that the ends are not more than 1 cm apart and at the same time do not touch each other. See what happens.

Observation: On letting the ends of tester dip into the lemon juice or vinegar, the bulb starts glowing.

Conclusion: This activity proves that both of the lemon juice and the vinegar are good conductors of electricity.

Liquids hich are Good Conductors: Some liquids which can conduct electricity are tap water lemon juice, vinegar, salt solution. Most of the liquids which conduct electricity are solutions of acids, bases or salts.

Liquids which are Poor Conductors: Some liquids are bad conductors of electricity, e.g.. distilled water, honey, milk. vegetable oil, etc.

Water: Tap water conducts electricity because it contains various salts dissolved in it. Hence, it is advised not to touch an electric switch with wet hands. Salt water also conducts electricity. However, distilled water is pure and free of salts, hence is a poor conductor.

Activity 2

Objective: To show that the distilled water is a poor conductor of electricity.
Materials Required: Distilled water, plastic or rubber cap.

Procedure:

• Take about two teaspoonful of distilled water in a clean and dry plastic or rubber cap of a bottle.
• Use the tester to test whether distilled water conducts electricity or not.
• Now dissolve a pinch of common salt in distilled water and test again. Record your observations.

Observation: The bulb of the tester does not glow when tester is put in distilled water. This means that distilled water does not conduct electricity. When tester is put in water dissolved with a pinch of salt, the bulb of the tester glows. This shows that when a pinch of salt is dissolved in distilled water, it conducts electricity.
Conclusion: Distilled water is a poor conductor of electricity.

Heating Effect of Electric Current:
The heating effect of electric current is responsible for the glowing of the bulb. When current passes through the bulb, the filament gets heated to a high temperature and as a result the bulb starts glowing. But, if current is very small, the filament will not get heated to a high temperature and will not glow.

Magnetic Effect of Electric Current:
When electric current passes through a conductor, a magnetic field is created around it. This is called magnetic effect of electric current. This can be demonstrated by keeping a magnetic compass near a current-carrying conductor. A magnetic compass shows deflection when it is placed near a current-carrying conductor. The deflection is the indicator of the presence of magnetic field.

Chemical Effects of Electric Current: A British Chemist named William Nicholson performed an experiment showing that if current is passed through water, bubbles of oxygen and hydrogen are produced. The oxygen bubbles will be present on the positive electrode and hydrogen bubbles on the negative electrode. The passage of an electric current through a conducting solution causes chemical reactions, due to which bubbles of a gas may be formed on the electrodes. Hence, it can be said that when electric current is passed through a conducting solution, some chemical reaction takes place in the solution. This is called the chemical effect of electric current. Some of the chemical effects of electric current are as follows:

• Bubbles of a gas may be formed on the electrodes.
• Deposits of metal may be seen on electrodes.
• Change of colour of solution may occur.

Potato tester: Current produces chemical effects in fruits and vegetables also. When the two ends of a tester are inserted inside a cut potato, the end connected to the positive terminal of the battery develops greenish-blue spot on the potato. This can be used to identify the positive terminal of a cell or battery concealed in a box.

Electroplating: The process of depositing a layer of a desired metal on any other material by means of electricity is called electroplating. Following steps are performed in electroplating:

• The material which needs to be coated with the desired layer is made the negative electrode (cathode).
• The plate of desired metal is made the positive electrode (anode).
• The conducting solution is made of a salt of desired metal.
• Electric current is passed through the solution.
• The desired metal dissociates from the plate (anode) and gets deposited on the material at negative electrode (cathode).

Electroplating of copper: During electroplating of copper, the copper plates are made the electrodes and copper sulphate solution is used. When electric current is passed through copper sulphate solution, copper sulphate dissociates into copper and sulphate. The free copper gets drawn to the negative electrode and gets deposited there. From the other electrode, an equal amount of copper gets dissolved in the solution and the process continues.

Activity 3

Objective: To show the process of electroplating.
Materials Required: Two copper plates of size around 10 cm x 4 cm, copper sulphate, distilled water, beaker, dilute sulphuric acid, sandpaper and battery.
Procedure:

• Take 250 mL of distilled water in a clean and dry vessel.
• Dissolve two teaspoonful of copper sulphate in it.
• Add a few drops of dilute sulphuric acid to the copper sulphate solution to make it more conducting.
• Clean the copper plates with sand paper.
• Now rinse them with water and dry them.
• Connect the copper plates to the terminals of battery and immerse them in copper sulphate solution.
• Allow the current to pass through the circuit for about 15 minutes.
• Now remove the electrodes from the solution and observe them carefully.

Observation: Copper metal gets deposited on the plate at negative terminal of the battery.

Conclusion: When the current is passed through the copper sulphate solution, copper gets deposited on the plate acting as cathode.

Applications of Electroplating:
i. Imitation jewellery is made by applying a layer of gold or silver.
ii. Applying a layer of chromium on an article by this method is called chrome plating. Parts of bicycle, motorbike and sanitary fittings are chrome plated by this method. Chromium is shiny, and resists corrosion and scratches.
iii. Tin cans are made by electroplating tin on iron.
iv. Applying a natural oxide layer on an article is called anodising. Electric poles and beams on bridges are electroplated with zinc. Chrome plating and anodising help in preventing articles from corrosion.

v. Galvanisation is a process in which a coating of zinc is deposited on iron to protect it from corrosion and formation of rust.

Problem of Electroplating: The disposal of the conducting solution of electroplating factories is a major problem. It is a polluting waste, and there are specific disposal guidelines which should be followed to protect the environment.

Class 8 Science Chapter 14 Chemical Effects of Electric Current Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Define good conductors of electricity.
Answer:
Electric current can easily pass through some materials. Such materials are called good conductors of electricity, e.g., iron, copper, silver, aluminium, gold, etc.

Question 2.
Define bad conductors of electricity.
Answer:
Electric current cannot pass easily through some materials. Such materials are called bad conductors of electricity, e.g., rubber, wood, asbestos, plastic, etc.

Question 3.
Why is zinc electroplated on iron?
Answer:
Zinc is electroplated on iron to prevent it from rusting.

Question 4.
Why is distilled water a poor conductor of electricity?
Answer:
Because distilled water is free from any kind of salts.

Question 5.
What effect does the current produce when it flows through a conducting solution?
Answer:
Current causes chemical reaction when it flows through a conducting solution.

Question 6.
What is galvanisation?
Answer:
Applying a layer of zinc on an iron article by electroplating is called galvanisation.

Question 7.
Why is it dangerous to touch an electrical appliance with wet hands?
Answer:
This is because water can conduct electricity.

Question 8.
What is LED?
Answer:
It is a semiconductor light source, used to detect a weak current in the circuit.

Question 9.
During electrolysis of water, oxygen gas is deposited at which electrode?
Answer:
Positive terminal or anode.

Question 10.
During electrolysis of water, hydrogen gas is deposited at which electrode?
Answer:
Negative terminal or cathode.

Question 11.
What is an electrolyte?
Answer:
A substance which conducts electricity in the liquid state or when dissolved in water and breaks up chemically during the process is called an electrolyte.

Question 12.
What is the practical use of LED?
Answer:
These are used in traffic lights and increasingly being used for lightning.

Question 13.
Define electric current.
Answer:
The amount of charge flowing per unit time is called electric current.

Question 14.
Name a liquid which does not conduct electricity.
Answer:
Distilled water.

Question 15.
What is the full form of LED?
Answer:
Light emitting diode.

Question 16.
Which of these do you think uses less electrical energy-electric bulb or LED?
Answer:
LED uses less electrical energy.

Short Answer Type Questions

Question 1.
Explain the heating effect of electric current.
Answer:
When an electric current passes through a conductor, it increases the temperature of the conductor. This is called heating effect of electric current. Many electrical appliances work on heating effect of electric current, e.g., electric bulb, water heater, electric iron, etc.

Question 2.
Explain the magnetic effect of electric current.
Answer:
When an electric current passes through a conductor, a magnetic field is created around the conductor. This is called magnetic effect of electric current. This can be demonstrated by keeping a magnetic compass near a current-carrying conductor. A magnetic compass shows deflection when it is placed near a current-carrying conductor. The devices which work on magnetic effect of electric current are electric motor and electromagnet.

Question 3.
What happens when electrodes are immersed in water and current is passed through it?
Answer:
When electrodes are immersed in water and current is passed through it, bubbles of oxygen and hydrogen gases are produced. Oxygen bubbles are formed at the electrode connected to the positive terminal. Hydrogen bubbles are formed at the electrode connected to the negative terminal.

Question 4.
Differentiate between conductors and insulators.
Answer:

Question 5.
Describe an electrical tester.
Answer:
An electrical tester is a simple piece of electronic test equipment used to determine the presence or absence of an electric voltage in a piece of equipment under test. It is also used to test whether a liquid allows electric current to pass through it or not.

Question 6.
Why is water that we get from ponds and hand pumps always a good conductor of electricity?
Answer:
Water that we get from ponds and hand pumps is not pure. It contains several amounts of mineral salts naturally dissolved in it, and thus is a good conductor.

Question 7.
Discuss what happens when an electrolyte is dissolved in water, and then an electric current is passed through it.
Answer:
When an electrolyte is dissolved in water, it breaks up into cations (positively charged ions) and anions (negatively charged ions). When an electric current is passed through the solution, the cations move towards the cathode (negatively charged electrode) and the anions move towards anode (positively charged electrode). This results in a chemical reaction or a chemical change.

Question 8.
What is the application of the chemical effects of electricity in our daily life? Give examples.
Answer:
Examples of chemical effect are as follows:

1. Electroplating: A layer of one metal is coated on the other substance or metal by the effect of electric current. This is called electroplating.
2. Electrolysis: The process by which a compound is decomposed into its constituents under the effect of electric current is called electrolysis.

Question 9.
List some of the chemical effects of electric current.
Answer:
When electric current is passed through a conducting solution, some chemical reaction takes place in the solution. This is called chemical effect of electric current. Some of the chemical effects of current are as follows:

• Bubbles of a gas may be formed at the electrodes.
• Deposits of metal may be seen on electrodes.
• Change of colour of solution may occur.

Question 10.
It is quite possible that even though a liquid allows electric current to pass through it and the circuit is complete but still the bulb does not glow? Why?
Answer:
When current passes through a substance, it offers opposition to the flow of current. If the opposition offered by the substance is small, the current will be large and if the opposition is large, the current will be small. However, if the current through the circuit is too small, the filament of the bulb does not get heated up to a sufficient temperature and it does not glow.

Long Answer Type Questions

Question 1.
Write the steps involved in electroplating.
Answer:
Following steps are involved in electroplating:

• The material which needs to be coated with a layer of desired metal is made the negative electrode (cathode).
• The plate of desired metal is made the positive electrode (anode).
• The conducting solution is made of a salt of desired metal.
• Electric current is passed through the solution.
• The electrolyte splits up releasing the ions of desired metal which get deposited at the cathode.
• An equal amount of the metal dissociates from the plate (anode) and gets dissolved in the electrolyte.

Question 2.
Write some uses of electroplating.
Answer:
Some uses of electroplating are as follows:

• Imitation jewellery is made by applying a layer of gold or silver.
• Applying a layer of chromium on an article by this method is called chrome plating. Parts of bicycle, motorbike and sanitary fittings are chrome plated by this method.
• Tin cans are made by electroplating tin on iron.
• Applying a layer of zinc on an article is called galvanisation.

Electric poles and beams on bridges are electroplated with zinc.

Question 3.
How will you show experimentally that an electric current can bring about a chemical change? What is this phenomenon called?
Answer:
Take two iron nails. Clean them with sandpaper. Wrap one or two rounds of copper wire around them and connect the other ends of the wires to the two terminals of an electric battery. Take water in a beaker and add to it a little salt or a few drops of sulphuric acid to make it conducting. Immerse the nails (called electrodes) in the solution. Observe the nails carefully.

You can see small bubbles of gases coming out from the water near the nails. It can be checked that the gases evolved are hydrogen and oxygen. The gases come from water-electric current breaks up water into its constituent gases, hydrogen and oxygen. This observation, therefore, shows that electric current has a chemical effect on water. This experiment shows that an electric current can bring about a chemical change. This phenomenon is called electrolysis.

Question 4.
Describe how you will prepare a tester to test conduction based on the magnetic effect of electricity.
Answer:
Take a tray of a discarded matchbox. Wrap an electric wire a few times around the tray. Place a smalt compass needle inside it. Now connect one free end of the wire to the terminal of a battery. Leave the other end free. Take another piece of wire and connect it to the other terminal of the battery. Join the free ends of the two wires momentarily. The compass needle should show deflection. The tester with two free ends of the wire is ready.

Question 5.
Complete the following table.

Answer:

Picture-Based Questions

Question 1.
Draw a well labelled diagram of passing current through water and answer the following questions.
a. Which material is used as electrodes?
b. Name the material used to make caps of electrodes.
c. Name the gases formed.
d. Can we call the change in water as a chemical change?
Answer:

a. Carbon rod is used as electrodes.
b. Brass caps or any other metal.
c. Oxygen and hydrogen.
d. Yes, it is a chemical change.

Question 2.
Observe the following circuits carefully. In which circuit will the bulb glow?

Answer:
a. No, because a rubber band is a poor conductor of electricity.
b. Yes, because iron key is a good conductor of electricity.
c. No, because pen is made of plastic which is a poor conductor of electricity.
d. Yes, because safety pin made of steel is a good conductor of electricity.