Module- 3C A (2023-2024)

Important Questions for Class 9 Mathematics

Chapter 13 – Surface Areas and Volumes

Very Short Answer Questions. 1 Mark

1. If the perimeter of one of the faces of a cube is 40 cm , them its volume is

(a) 6000 cm3

(b) 1600 cm3
(c) 1000 cm3
(d) 600 cm3
Ans: (c) 1000 cm3

The side of the one face of cube = 40 cm = 4a

40 cm
a= = 10 cm
4

Volume of the cube is V = a3

V = aaa

=101010

= 1000 cm3

2. A cuboid having surface areas of 3 adjacent faces as a, b and c has the

volume

(a) 3

(b)

Module- 3C A (2023-2024)
 (c) abc
(d) a3b3c3

Ans: (b)

Let length, width and height of cuboid be w and h respectively

Considering adjacent faces: AEHD, DHGC and EFGH

Let area of AEHD = a , area of DHGC = b and area of EFGH = c

Also, area of AEHD = lw

Area of DHGC = wh

Area of EFGH = lh

Therefore, lw = a, wh = b and lh = c

 lw whlh = a bc

 l2w2h2 = abc

 (lwh)2 = abc

V 2 = abc

V =

Volume of cuboid is V =

3. The diameter of a right circular cylinder is 21 cm and its height is 8 cm. The
Volume of the cylinder is

(a) 528 cm3

(b) 1056 cm3
(c) 1386 cm3
(d) 2772 cm3

Module- 3C A (2023-2024)
 Ans: (d) 2772 cm3

Diameter of Cylinder = 21 cm.

Height = 8 cm.
D
Radius of Sphere =
2

Volume of Cylinder = ( r 2 h )

21 2
V = ( )( ) 8
2
 21 2
=     8
 2 

= 2770.88472 cm3

V = 2772 cm3

Volume of right circular cylinder is

4. Each edge of a cube is increased by 40% . The % increase in the surface area
is.

(a) 40

(b) 96

(c) 160

(d) 240
Ans: (b) 96

Let the edge of the cube be equal to 'a' units.

Thus, the initial surface area ( A1) = a2units2

Now, the edge of the cube increases by 40%

Module- 3C A (2023-2024)
 The new edge length = a + 40% of a = 1.4a .

Thus, the final surface area (A 2 ) = (1.4a)2 = 1.96a2 units 2

(1.96a2 − a2 )
Percentage change = [( A2 − A1 ) / ( A1 )]100 =  100
 (a2 ) 
= 0.96100

= 96%

5. Find the curved (lateral) surface area of each of the following right circular
cylinders:

(a) 2 rh
(b)  rh
(c) 2r(r + h)

(d) None of these

Ans: (a) 2 rh

Lateral Surface Area or Curved Surface Area of a Right Circular Cylinder

= (Perimeter of the Cross Section) × Height

Module- 3C A (2023-2024)
 = 2πrh

6. The radius and height of a right circular cylinder are each increased by
20% . The volume of cylinder is increased by-

(a) 20%

(b) 40%

(c) 54%

(d) 72.8%

Ans: (d) 72.8%

Volume = r2 h
20
new radius = r + r
100
5
= r
6

So
20 6
=h+ h= h
100 5
 6r 2  6 
Volume =     h
 5  5 
216
=  r2 h
125

 Increase in
216
Volume =  r2 h
125

= 72.8%

Module- 3C A (2023-2024)
 7. A well of diameter 8 meters has been dug to the depth of 21 m . the volume
of the earth dug out is
(a) 1056cu m

(b) 352cum

(c) 1408cum

(d) 4224cum

Ans: (a) 1056m3

Volume of the well is V = r2h

V =   4 4 21

=1056m3

8. The radius of a cylinder is doubled and the height remains the same. The
ratio between the volumes of the new cylinder and the original cylinder is
(a) 1: 2

(b) 1: 3

(c) 1: 4

(d) 1: 8

Ans: (c) 1: 4

The radius of a cylinder is doubled and the height remains the same. (Given)

Radius of original cylinder = r

Radius of new cylinder = 2r

Height remains the same.

Module- 3C A (2023-2024)
 We know that,

Volume of new cylinder =  (2r)2 h

Volume of new cylinder = 4r2h

Now

Let ratio of volume be " x ".

Ratio of volume = Volume of new cylinder / Volume of original cylinder

[ Put the values]

x = 4r2h / r2h

x = 4r2 / r2

x = 4 /1

The ratio between the volumes of the new cylinder and original cylinder is \[1:4.\]

9. Length of diagonals of a cube of side a cm is

(i) 2a cm

(ii) 3a cm

(iii) 3a cm

(iv) 1cm

Ans: (ii) 3a cm

Diagonal of a Cube = 3x

Where is the cube side.

10. Surface area of sphere of diameter 14 cm is

Module- 3C A (2023-2024)
 (i) 616 cm2

(ii) 516 cm2

(iii) 400Cm2
(iv) 2244 cm2

Ans: (i) 616 cm2

Given Diameter of sphere = 14 cm radius = 7 cm

Surface area of sphere = 4 r2 = 4 (7)2 = 43.14 49

Surface area of sphere = 616 cm2

11. Surface area of bowl of radius r cm is

(i) 4r2

(ii) 2r2

(iii) 3 r2
(iv) r2
Ans: (iii) 3 r2

The area of a circle of radius r is r2

Thus if the hemisphere is meant to include the base then the surface area is
2 r2 +  r2 = 3 r2

12. Volume of a sphere whose radius 7 cm is

(i) 1437 1 cm3

3
1 3
(ii) 1337 cm
3

Module- 3C A (2023-2024)
 (iii) 1430 cm3

(iv) 1447 cm3

Ans: (i) 1437 1 cm3

3

Radius = 7 cm

4
Volume of sphere =  r
3

3
 4 22 
=   7  7  7 cm3
3 7 
 
4 
=  221 7  7 cm3
3 
 

4312 3
= cm
3

= 1437.33 cm3

13. The curved surface area of a right circular cylinder of height 14 cm is

88 cm2 . find the diameter of the base of the cylinder

(i) 1 cm

(ii) 2 cm

(iii) 3 cm

(iv) 4 cm

Ans: (ii) 2 cm

Given, The height of cylinder = 14 cm and, the curved surface area of cylinder
= 88 cm2

The curved surface area of cylinder = 2 rh and 2r = d

Module- 3C A (2023-2024)
 Here, r = radius of cylinder, d = diameter of cylinder and

h = height of cylinder

So, the curved surface area of cylinder =  dh = 88 cm2

 d14 = 88

3.14 d 14 = 88

d = 2 cm

So, the diameter of the cylinder is 2 cm .

14. Volume of spherical shell

2
(i)  r3
3
3
(ii)  r3
4

4
(iii)  R3 − r3 
3

(iv) none of these

4
Ans: (iii)   R3 − r3 
3 
4
Volume of outer sphere =  R
3

3
4
Volume of inner sphere =  r
3

Total net volume between both the spheres =  (R3 − r3 )

4
3

Module- 3C A (2023-2024)
 15. The area of the three adjacent faces of a cuboid are x, y, z . Its volume is V ,
then

(i) V = xVZ
(ii) V 2 = xyz

(iii) V = x2 y2z2

(iv) none of these

Ans: (ii) V 2 = xyz

Let the 3 dimensions of the cuboid be l, b and h so,

x = lb

y = bh

z = hl

Multiplying above three equations,

xyz = lbbhhl

= 12 b2 h2

As,

V = lbh

So,

V2 = l2 b2 h2

V2 = xyz

16. A conical tent is 10 m high and the radius of its base is 24 m then slant
height of the tent is

(i) 26

Module- 3C A (2023-2024)
 (ii) 27

(iii) 28

(iv) 29
Ans: (i) 26

Height (h) of conical tent = 10 m

Radius (r) of conical tent = 24 m

Let the slant height of the tent be l

l2 = h2 + r2

l2 = (10)2 + (24)2

l2 = 100 + 576

l2 = 676

l=

l = 262

l = 26 m

Therefore, the slant height of the tent is 26 m .

17. Volume of hollow cylinder

(i)  ( R 2 − r 2 )h

(ii) R2h

(iii) r2h

(iv)  r 2 ( h1 − h2 )

Ans: (i)  ( R 2 − r 2 )h

Module- 3C A (2023-2024)
 The formula to calculate the volume of a hollow cylinder is given as,

Volume of hollow cylinder =  (R2 − r2 )h cubic units,

where, R is the outer radius, ' r ' is the inner radius, and, ' h ' is the height of the
hollow cylinder.

18. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm . then
curved surface area.

(i) 155 cm2

(ii) 165 cm2

(iii) 150 cm2

(iv) none of these

Ans: (ii) 165 cm2

Diameter of the base of the cone is 10.5 cm and slant height is 10 cm .

Curved surface area of a right circular cone of base radius, \['r'\]and slant height,

is  r .

Diameter, d =10.5 cm

Radius, r =10.5 / 2 cm = 5.25 cm

Slant height, l =10 cm

Curved surface area =  rl

= 3.145.2510 =165 cm2

Thus, curved surface area of the cone =165 cm2 .

Module- 3C A (2023-2024)
 19. The surface area of a sphere of radius 5.6 cm is

(i) 96.8cm2

(ii) 94.08cm2
(iii) 90.08cm2
(iv) none of these

Ans: (ii) 94.08cm2

Given radius of sphere = 5.6 cm

Surface area of sphere = 4 r2

= 43.14(5.6)2

Surface area of sphere = 393.88 cm2

20. The height and the slant height of a cone are 21 cm and 28 cm respectively
then volume of cone

(i) 7556 cm3

(ii) 7646 cm3

(iii) 7546 cm3

(iv) None of these

Ans: (c) 7546 cm3

1
Volume of the cone =  r h
2

Given

Slant height = l = 28 cm

Height of cone = h = 21 cm

Module- 3C A (2023-2024)
 Let radius of cone = r cm

l2 = h2 + r2

282 = 212 + r2

282 − 212 = r2

r2 = 282 − 212

r2 = (28 − 21)(28 + 21)

r 2 = (7)(49)

r=

r = 7(7)2

r = 7 7 cm

1
Volume of the cone =  r h
2

3
1 22
=   7 7  7 7  21 cm3
3 7

= 22  7 7  7 7 cm3

= 22 7  7 ( 7 )2 cm3

= 22 7 7 7 cm3

= 7546 cm3

2. The length, breadth and height of a room are 5 m, 4 m and 3m respectively.

Find the cost of white washing the walls of the room and the ceiling at the rate
of Rs. 7.50 per m2

Ans. Given: Length (l) = 5 m , Breadth (b) = 4 m and Height (h) = 3m

Module- 3C A (2023-2024)
 Area of the four walls = Lateral surface area = 2(bh + hl) = 2h(b + l)

= 23(4 + 5)

= 293 = 54 m2

Area of ceiling = l b = 5 4 = 20 m2

 Total area of walls and ceiling of the room = 54 + 20 = 74 m2

Now cost of white washing for 1m2 = Rs. 7.50

 Cost of white washing for 74 m2 = 747.50 = Rs. 555

3. The floor of a rectangular hall has a perimeter 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.

Ans. Given: Perimeter of rectangular wall = 2(l + b) = 250 m ............. (i)

Now Area of the four walls of the room

Total cost to paint walls of the room

=
Cost to paint 1m2 of the walls

15000
= = 1500 m2  (ii)
10

Area of the four walls = Lateral surface area = 2(bh + hl) = 2h(b + l) =1500

 250 h =1500

1500
h= = 6m
250

Hence required height of the hall is 6 m.

Module- 3C A (2023-2024)
 4. The paint in a certain container is sufficient to paint an area equal to
9.375 m2 . How many bricks of dimensions 22.5cm10cm7.5cm can be painted
out of this container?

Ans. Given: Length of the brick (l) = 22.5cm, Breadth (b) =10cm and Height
(h) = 7.5m

 Surface area of the brick = 2(lb + bh + hl)

= 2(22.510 +107.5 + 7.5 22.5)

= 2(225 + 75 + 468.75)

= 937.5cm2

= 0.09375m2

Now No. of bricks to be painted

Total area to be painted

=
Area of one brick

9.375
= = 100
0.09375

Hence 100 bricks can be painted.

5. A cubical box has each edge 10cm and a cuboidal box is 10cm wide, 12.5cm
long and 8cm high.

(i) Which box has the greater lateral surface area and by how much?

Ans. (i) Lateral surface area of a cube = 4( side )2 = 4(10)2 = 400cm2

Lateral surface area of a cuboid = 2h(l + b) = 28(12.5 +10)

=16 22.5 = 360cm2

Module- 3C A (2023-2024)
  Lateral surface area of cubical box is greater by (400 − 360) = 40cm2

(ii) Which box has the smaller total surface area and how much?

(ii) Total surface area of a cube = 6( side )2 = 6(10)2 = 600cm2

Total surface area of cuboid = 2(lb + bh + hl) = 2(12.510 +108 + 812.5)

= 2(125 + 80 +100)

= 2305 = 610cm2

 Total surface area of cuboid box is greater by (610 − 600) = 10cm2

6. Praveen wanted to make a temporary shelter for her car, by making a box-
like structure with tarpaulin that covers all the 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 much tarpaulin
would be required to make the shelter of height 2.5m with base dimensions
4 m3m ?

Ans. Given: Length of base (l) = 4 m , Breadth (b) = 3m and Height (h) = 2.5m

Tarpaulin required to make shelter = Surface area of 4 walls + Area of roof

= 2h(l + b) + lb = 2(4 + 3)2.5 + 43

= 35 +12 = 47m2

Hence 47 m2 of the tarpaulin is required to make the shelter for the car.

7. The curved surface area of a right circular cylinder of height 14cm is 88cm2.
Find the diameter of the base of the cylinder.

Module- 3C A (2023-2024)
 Ans. Given: Height of cylinder (h) =14cm , Curved Surface Area = 88cm2

Let radius of base of right circular cylinder = r cm

2 rh = 88

22
 2  r 14 = 88
7

 r = 88 7  1  1
22 14 2

 r =1cm

Diameter of the base of the cylinder = 2r = 21 = 2cm

8. It is required to make a closed cylindrical tank of height 1m and base

diameter 140 cm from a metal sheet. How many square meters of the sheet are
required for the same?

Module- 3C A (2023-2024)
 Ans. Given: Diameter =140cm

 Radius (r) = 70cm = 0.7 m

Height of the cylinder (h) = 1m

22
Total surface Area of the cylinder = 2r(r + h) = 2  0.7(0.7 +1)
7

= 2 220.11.7 = 7.48m2

Hence 7.48m2 metal sheet is required to make the close cylindrical tank.

9. The diameter of a roller is 84cm and its length is 120cm . It takes 500
complete revolutions to move once over to level a playground. Find the area of
the playground in m2

Ans. Diameter of roller = 84 cm

 Radius of the roller = 42 cm

Length (Height) of the roller =120cm

Curved surface area of the roller = 2 rh

22
= 2  42120 = 31680 cm2
7

Now area leveled by roller in one revolution = 31680cm2

Module- 3C A (2023-2024)
  Area leveled by roller in 500 revolutions = 3.1680500 =1584.0000

=1584 m2

10. A cylindrical pillar is 50cm in diameter and 3.5m in height. Find the cost of
white washing the curved surface of the pillar at the rate of Rs. $12.50$ per
m2

Ans. Diameter of pillar = 50cm

 Radius of pillar = 25 cm = 25 = 1 m
100 4

Height of the pillar = 3.5 m

22 1
Now, Curved surface area of the pillar = 2rh = 2   3.5
7 4

11 2
= m
2

Cost of white washing 1m2 = Rs.12.50

11 11
 Cost of white washing m2 = 12.50
2 2

= Rs.68.75

11. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of
the base of the cylinder is 0.7 m , find its height.

Ans. Curved surface area of the cylinder = 4.4 m2 ,

Radius of cylinder = 0.7 m

Let height of the cylinder = h

Module- 3C A (2023-2024)
 2 rh = 4.4

22
 2  0.7  h = 4.4
7

h = 4.4 7  1  1
22 2

 h =1m

12. The inner diameter of a circular well is 3.5m . It is 10 m deep. Find:

(i) its inner curved surface area.

Ans. Inner diameter of circular well = 3.5m

3.5
 Inner radius of circular well = = 1.75 m
2

And Depth of the well = 10 m

(i) Inner surface area of the well = 2 rh

22
= 2 1.7510 = 110 m2
7

(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2 .

Ans. Cost of plastering 1m2 = Rs. 40

Module- 3C A (2023-2024)
 Cost of plastering 100 m2 = 40110 = Rs. 4400

13. In a hot water heating system, there is a cylindrical piping of length 28m
and diameter 5cm . Find the total radiating surface in the system.

Ans. The length (height) of the cylindrical pipe = 28 m

Diameter = 5cm
5
 Radius = cm
2

22 5
Curved surface area of the pipe = 2 rh = 2   2800
7 2

44000
= 44000 cm2 = = 4.4 m2
10000

14. In the adjoining figure, you see the frame of a lampshade. It is to be

covered with a decorative cloth. The frame has a base diameter of 20cm and
height of 30cm . A margin of 2.5cm is to be given for folding it over the top and
bottom of the frame. Find how much cloth is required for covering the
lampshade.

Ans. Height of each of the folding at the top and bottom (h) = 2.5cm

Height of the frame (H) = 30cm

Diameter = 20 cm

 Radius = 10 cm

Now cloth required for covering the lampshade

=CSA of top part + CSAof middle part +CSA of bottompart

Module- 3C A (2023-2024)
 = 2rh + 2rH + 2rh

= 2r(h + H + h)

= 2r(H + 2h)

22
=2 10(30 + 2 2.5)
7

= 2200cm2

15. The students of a Vidyalaya were asked to participate in a competition for

making and decorating penholders in the shape of a cylinder with a base,
using cardboard. Each penholder was to be of radius 3cm and height 10.5cm.
The Vidyalaya was to supply the competitors with cardboard. If there were 35
competitors, how much cardboard was required to be bought for the
competition?

Ans. Radius of a cylindrical pen holder (r) = 3cm

Height of the cylindrical pen holder (h) =10.5cm

Cardboard required for pen holder = CSA of pen holder + Area of circular base

= 2rh +  r 2 = r(2h + r)

22
=  3(210.5 + 3) = 226.28 cm2
7

Since Cardboard required for making 1 pen holder = 226.28cm2

 Cardboard required for making 35 pen holders = 226.2835 = 7919.8cm2

= 7920cm2 (approx.)

Module- 3C A (2023-2024)
 16. Diameter of the base of a cone is 10.5cm and its slant height is 10cm . Find
its curved surface area and its total surface area.

Ans. Diameter =10.5cm

10.5 21
 Radius (r) = = cm
2 4

Slant height of cone (l) =10cm

22 21
Curved surface area of cone =  rl =  10
7 4

=165cm2
22 21  21 
Total surface area of cone =  r(l + r) = 7  4 10 + 4 
 

22 21 61
=   = 251.625 cm2
7 4 4

Module- 3C A (2023-2024)
 17. Find the total surface area of a cone, if its slant height is 21cm and
diameter of the base is 24cm .

Ans. Slant height of cone (l) = 21m

Diameter of cone = 24 m
24
 Radius of cone (r) = = 12 m
2

Total surface area of cone = r(l + r)

22
= 12(21+12)
7

264
=  33 = 1244.57 m2
7

18. The slant height and base diameter of a conical tomb are 25m and 14 m
respectively. Find the cost of whitewashing its curved surface at the rate of Rs.
210 per 100 m2

Module- 3C A (2023-2024)
 Ans. Slant height of conical tomb (l) = 25m, Diameter of tomb = 14 m

14
 Radius of the tomb (r) = = 7m
2

22
Curved surface are of tomb =  rl =  7  25 = 550 m2
7

Cost of white washing 100 m2 = Rs. 210

210
 Cost of white washing 1m2 =
100

210
 Cost of white washing 550 m =  550
2

100

= Rs. 1155

19. A Joker's cap is in the form of a right circular cone of base radius 7 cm and
height 24 cm . Find the area of the sheet required to make 10 such caps.

Ans. Radius of cap (r) = 7 cm, Height of cap (h) = 24cm

Slant height of the cone (l) = r 2 + h2 = (7)2 + (24)2

Module- 3C A (2023-2024)
 = = = 25cm

Area of sheet required to make a cap = CSA of cone =  rl

22
=  7  25 = 550 cm2
7

 Area of sheet required to make 10 caps =10550 = 5500cm2

20. Find the surface area of a sphere of radius:

(i) 10.5cm

Ans. Radius of sphere =105cm

22
Surface area of sphere = 4  r 2 = 4 10.510.5
7

=1386cm2

(ii) 5.6cm

Ans. Radius of sphere = 5.6 m

22
Surface area of sphere = 4  r = 4  5.6 5.6
2

= 3.94.84 m2

(iii) 14cm

Ans. Radius of sphere = 14 cm

22
Surface area of sphere = 4  r = 4 1414
2

= 2464cm2

Module- 3C A (2023-2024)
 21. Find the surface area of a sphere of diameter:

(i) 14cm

Ans. (i) Diameter of sphere = 14 cm ,

14
Therefore, Radius of sphere = = 7 cm
2

22
Surface area of sphere = 4  r = 4  7  7 = 616 cm2
2

(ii) 21cm

Ans. Diameter of sphere = 21cm

21
 Radius of sphere = cm
2

22 21 21
Surface area of sphere = 4  r = 4  
2

7 2 2

=1386cm2

(iii) 3.5cm

Ans. Diameter of sphere = 3.5cm

3.5
 Radius of sphere = = 1.75 cm
2

22
Surface area of sphere = 4  r 2 = 4 1.751.75
7

= 38.5cm2

22. Find the total surface area of a hemisphere of radius 10cm . (Use  = 3.14 )

Module- 3C A (2023-2024)
 Ans. Radius of hemisphere (r) =10cm

Total surface area of hemisphere = 3 r2

= 33.141010

= 942cm2

Hence total surface area of hemisphere is 942 cm2 .

23. Find the radius of a sphere whose surface area is 154 cm2 .

Ans. Surface area of sphere =154cm2

 4r2 =154

22
 4  r 2 = 154
7

154  7
 r2 =
22  4

49
 r2 =
4

7
r= = 3.5 cm
2
Module- 3C A (2023-2024)
 24. A hemispherical bowl is made of steel, 0.25cm thick. The inner radius of
the bowl is 5cm . Find the outer curved surface area of the bowl.

Ans. Inner radius of bowl (r) = 5cm

Thickness of steel (t) = 0.25cm

 Outer radius of bowl (R) = r + t = 5 + 0.25 = 5.25cm

22
 Outer curved surface area of bowl = 2 R2 = 2  5.25 5.25
7

22 21 21
= 2  
7 4 4

693
= = 173.25 cm2
4

25. A right circular cylinder just encloses a sphere of radius r (See figure).
Find:

(i) Surface area of the sphere.

Ans. Radius of sphere = r

 Surface area of sphere = 2 ( radius )2 = 2r2

(ii) Curved surface area of the cylinder.

Ans. The cylinder just encloses the sphere in it.

 The height of cylinder will be equal to diameter of sphere.

And The radius of cylinder will be equal to radius of sphere.

 Curved surface area of cylinder = 2rh = 2r r

Module- 3C A (2023-2024)
 = 4r2

(iii) Ratio of the areas obtained in (i) and (ii).

Ans. Surface area of sphere 4r2 1

= =
Curved surface area of cylinder 4  r2 1

 Required ratio =1:1

26. A matchbox 4cm 2.5cm1.5cm . What will be the volume a packet

containing 12 such boxes?

Ans. Given: Length (l) = 4cm ,

Breadth (b) = 2.5cm ,

Height (h) =1.5cm

Volume of a matchbox = l b h

= 4 2.51.5

= 15cm3

Module- 3C A (2023-2024)
  Volume of a packet containing 12 such matchboxes is 180 cm3.

27. A cubical water tank is 6 m long, 5m wide and 4.5m deep. How many litres
of water can it hold?

Ans. Here l = 6 m,b = 5m and h = 4.5m

 Volume of the tank = l b h

= (65 4.5)cm3

= 135 m3

=1351m3

=1351000 litres

=135000 litres

So, the cuboidal water tank can hold 135000 litres of water.

28. A cuboidal vessel is 10 m long and 8m wide. How high must it be to hold
380 cubic meters of a liquid?

Ans. Let height of cuboidal vessel = h m

Length = 10 m

Breadth = 8m

Volume of liquid in cuboidal vessel = 380 m3

 l b h = 380 m3

10m8m h = 380

Module- 3C A (2023-2024)
 380
h= = 4.75 m
108

Hence cuboidal vessel is 4.75 m high.

29. Find the cost of digging a cuboidal pit 8m long. 6 m broad and 3m deep at
the rate of Rs. 30 perm3 .

Ans. Here, l = 8m,b = 6 m and h = 3m

Volume of the cuboidal pit = lbh

= (8 63)m3

= 144 m3

Cost of digging 1m3 = Rs 30

Cost of digging 144 m3 = Rs(14430)

= Rs 4320

Cost of digging the pit is Rs 4320

30. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth
of the tank, if its length and depth are respectively 2.5m and 10 m.

Ans. Length = 2.5m

Height = 10 m

Let Breadth be b m

Capacity of cuboidal tank = 50000 liters

 l b h = 50000 liters

Module- 3C A (2023-2024)
 50000
 2.5 m b 10 m = m3
1000

 25b = 50

 b = 2m

Hence breadth of cuboidal tank is 2 m .

31. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour.
How much water eill fall into the sea in a minute?

Ans. Water flowing in river in 1 hour = 2 km

Water flowing in river in 1 hour = 2000 m

Water flowing in river in 60 minutes = 2000 m

2000 100
Water flowing in river in 1 minute = m= m
60 3

Now,

River is in shape of cuboid

100
Length = m
3

Breadth = 40 m

Height = 3m

Volume of water falling in the sea in 1 minute = Volume of the cuboid

=Length  Breadth  Height

 100 
=  40 3

 3 3 m
 

= (100  40 1)m3

Module- 3C A (2023-2024)
 = 4000 m3

32. Find the length of a wooden plank of width 2.5m , thickness 0.025 m and
volume 0.25m3

Ans. Given: Volume of wooden plank = 0.25m3

 l  2.50.025 = 0.25

0.25
l =
2.5 0.025

 l = 4m

Hence required length of wooden plank is 4 m .

33. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5cm , then (i)
radius of its base

Ans. Let radius of cylinder = r cm

Height = h = 5cm

Now it is given that

Lateral surface = 94.2cm2

Curved surface area of cylinder = 94.2cm2

2 rh = 94.2

23.14 r 5 = 94.2

94.2
r=
2  3.14  5

r = 3cm

Module- 3C A (2023-2024)
 (ii) volume of the cylinder.

Ans. r = 3cm,

h = 5cm

Volume of cylinder =  r 2 h

= 3.14335

=141.3cm3

34. A bag of grain contains 2.8 m3 of grain. How many bags are needed to fill a
drum of radius 4.2m and height 5m?

Ans. Given

Volume of grain inside the bag = 2.8 m3

Radius of the drum = 4.2 m

Height of the drum = 5m

 Volume of the drum =  r 2 h

22
= (2.1)2  5
7

The number of bag full of grains required

Volume of the drum

=
Volume of the bag

22
 2.1 2.1 5
= 7 = 99 bags
2.8

Hence 99 bags are needed to fill the drum.

Module- 3C A (2023-2024)
 35. A lead pencil consists of a cylinder of wood with a solid cylinder of
graphite filled in the interior. The diameter of the pencil is 7 mm and diameter
of graphite is 1mm . If the length of the pencil is 14cm , find the columns of the
wood and that of the graphite.

Ans. Diameter of graphite = 1mm

22
Volume of graphite =  r 2 h = (0.05)2 14 = 0.11cm3
7

Diameter of pencil = 7 mm

 Radius of pencil (R) = 3.5mm = 0.35cm

22
Volume of pencil =  R 2 h = (0.35)2 14 = 5.39 cm3
7

Now, Volume of wood = Volume of pencil −Volume of graphite

= 5.39 − 0.11 = 5.28cm3

36. A patient in a hospital is given soup daily in a cylindrical bowl of diameter

7cm . If the bowl is filled with soup to a height of 4cm , how much soup the
hospital has to prepare daily to serve 250 patients?

Ans. Soup is in form of cylinder with

7
Radius = r = Diameter = cm
2 2

Height = h = 4cm

Volume of the soup in cylindrical bow =  r 2 h

22 7 7
=    4 cm3
7 2 2

Module- 3C A (2023-2024)
 =154cm3

Soup served to 1 patient =154cm3

Soup served to 250 patients = 250154cm3

= 38500cm3

1
= 38500  litres
1000

= 38.5 litres

37. Find the volume of the right circular cone with:

(i) Radius 6cm , Height 7 cm

Ans. Given: r = 6cm, h = 7 cm

1
Volume of cone =  r h
2

1 22
=   6 6 7
3 7

= 264cm3

(ii) Radius 3.5cm , Height 12cm

Ans. Given: r = 3.5cm, h =12cm

1
Volume of cone =  r h
2

1 22
=   3.5 3.512
3 7

=154cm3

Module- 3C A (2023-2024)
 38. The height of a cone is 15cm . If its volume is 1570 cm3 , find the radius of the
base.

Ans. Height of cone = h =15cm

Let radius of cone = rcm

Given

Volume of cone =1570cm3

1
 3.14 r 2 15 = 1570
3

13.14 r2 5 =1570

1570
r2 =
3.14 5

r2 =100r

= 100r = (10)2

r =10cm

Hence required radius of the base is 10 cm.

39. If the volume of a right circular cone of height 9cm is 48 cm3 , find the
diameter of the base.

Ans. Height of the cone (h) = 9cm

Let radius of cone = rcm

Given Volume of cone = 48cm3

Module- 3C A (2023-2024)
 1
 r 2 h = 48
3

1
  r 2 h = 48
3

1
  r 2  9 = 48
3

 3r2 = 48

48
 r2 = = 16
3

 r = 4cm

 Diameter of base = 2r = 2 4 = 8cm

40. A conical pit of top diameter 3.5m is 12 m deep. What is its capacity in
kiloliters?

Ans. Height of conical pit = h =12 m

3.5
Radius of conical pit = r = Diameter = m = 1.75 m
2 2

1 2
Capacity of pit = Volume of cone =  r h
3

 1 22
= 3  7 1.751.75 12 m 3

 

= 38.5m3

= 38.5 kiloliters

Capacity of pit = 38.5 kiloliters.

Module- 3C A (2023-2024)
 41. A right triangle ABC with sides 5cm,12cm and 13cm is revolved about the
side 12 cm . Find the volume of the solid so obtained. (Use  = 3.14 )

Ans. When right angled triangle ABC is revolved about side 12cm, then the solid
formed is a cone.

In that cone, Height (h) =12cm

And radius (r) = 5cm

1
Therefore, Volume of cone =  r h
2

1
=   5 512
3

=100cm3

42. Find the volume of the largest right circular cone that can be fitted in a
cube whose edge is 14cm .
1
Ans. For largest circular cone radius of the base of the cone = edge of cube
2

1
= 14 = 7cm
2

And height of the cone = 14 cm

1
Volume of cone =  3.14 7  7 14
3

= 718.666cm3

43. Find the volume of a sphere whose radius is

(i) 7 cm
Module- 3C A (2023-2024)
 Ans. Radius of sphere (r) = 7 cm
4
Volume of sphere =  r
3

4 22
=  777
3 7

4312 1437 1 cm3

= =
3 3

(ii) 0.63cm

Ans. Radius of sphere (r) = 0.63m

4 3
Volume of sphere =  r
3

4 22
=   0.63 0.63 0.63
3 7

4 22 63 63 63
=    
3 7 100 100 100

= 1.047816m3 =1.05m3 (approx.)

44. Find the amount of water displaced by a solid spherical ball of diameter:

(i) 28cm

Ans. Diameter of spherical ball = 28cm

28
 Radius of spherical ball (r) = = 14 cm
2

According to question, Volume of water replaced = Volume of spherical ball

4
= r3
3

Module- 3C A (2023-2024)
 4 22
=  141414
3 7

34496 11498 2 cm3

= =
3 3

(ii) 0.21m

Ans. Diameter of spherical ball = 0.21m

0.21
 Radius of spherical ball (r) = m
2

According to question,
4
Volume of water replaced = Volume of spherical ball =  r
3

4 22 0.21 0.21 0.21

=    
3 7 2 2 2

4 22 21 21 21
=    
3 7 200 200 200

441
= 11 = 0.004851m3
100 100 100

45. The diameter of a metallic ball is 4.2cm . What is the mass of the ball, if the
metal weighs 8.9g per cm3 ?

Ans. Diameter of metallic ball = 4.2 cm

4.2
 Radius of metallic ball (r) = = 2.1cm
2

4
Volume of metallic ball =  r
3

Module- 3C A (2023-2024)
 4 22
=   2.1 2.1 2.1
3 7

4 22 21 21 21
=     = 38.808 cm3
3 7 10 10 10

Density of metal = 8.9 g per cm3

Mass of 38.808cm3 = 8.938.808

= 345.3912g = 345.39g

46. A hemispherical tank is made up of an iron sheet 1cm thick. If the inner
radius is 1 m , then find the volume of the iron used to make the tank.

Ans. Inner radius = r1 = 1m

Outer radius = r2 =1m+1cm

1
= 1m + m
100

=1m + 0.01m

= 1.01m

Volume of iron used =Volume of outer hemisphere − Volume of inner hemisphere

2
Volume of iron of hemisphere =  R3 − r3 
3

2 22
=   (101)3 − (100)3 
3 7  

44
= [1030301−1000000]
21

= 0.06348m3

Module- 3C A (2023-2024)
 47. A dome of a building is in the form of a hemisphere. From inside, it was
whitewashed at the cost of Rs. 498.96.If the cost of white-washing is at the rate
of Rs. 2.00 per square meter, find:

(i) the inner surface area of the dome.

Ans. Cost of white washing from inside = Rs.498.96

Rate of white washing = Rs.2

498.96
Area white washed = = 249.48 cm2
2

Therefore, inner surface area of dome = 249.48m2

(ii) the volume of the air inside the dome.

2
Ans. Volume of air inside dome = Volume of hemisphere =  r 3
3

Let the radius of dome = r m

First we find radius using surface area

Surface area of dome = 249.48m2

2r2 = 249.48

22
2  r 2 = 249.48
7

249.48 7
r2 =
2  22

r2 = 39.69

r=

r = 6.3m

Module- 3C A (2023-2024)
 Volume of the air inside the dome = volume of hemisphere
2
= r3
3

2 22
=   6.3 6.3 6.3 m3
3 7

= 523.908m3

48. Twenty-seven solid iron spheres, each of radius r and surface area S are
melted to form a sphere with surface area S '. Find the:

(i) radius r ' of the new sphere.

4
Ans. Volume of 1 sphere, V =  r 3
3

Volume of 27 solid sphere

4
= 27   r 3
3

Let r1 is the radius of the new sphere.

Volume of new sphere = Volume of 27 solid sphere

4 4
 r 3 = 27   r 3
3 1 3

r13 = 27
r3

r1 3
= 27
r

Module- 3C A (2023-2024)
 r1 3
=
r 1

r1 = 3r

(ii) ratio of S and S '.

Surface area of new sphere S1

Ans.
Surface area of old sphere S

S 4r2
1
= 1

S 4r2

S1 (3r)2
=
S r2

S1 (3r)2
=
S r2
2
S1 9r
= 2
S r

S1 9
=
S 1

S1 : s = 9 :1

s : S1 =1: 9

49. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm . How

much medicine (in mm3 ) is needed to fill this capsule?

Ans. Diameter of spherical capsule = 3.5 mm

3.5 35 7
 Radius of spherical capsule (r) = = = mm
2 20 4

Medicine needed to fill the capsule = Volume of sphere

Module- 3C A (2023-2024)
 4 4 22 7 7 7
= r3 =    
3 3 7 4 4 4

11 7  7 539
= = mm3
3 2 4 34

= 22.46 mm3

50. Sameera wants to celebrate the fifth birthday of her daughter with a
party. She bought thick paper to make the conical party caps. Each cap is to
have a base diameter of 10cm and height 12cm . A sheet of the paper is 25cm by
40cm and approximately 82% of the sheet can be effectively used for making
the caps after cutting. What is the minimum number of sheets of paper that
Sameera would need to buy, if there are to be 15 children at the party? (Use
 = 3.14)

Ans. Diameter of base of conical cap = 10cm

 Radius of conical cap (r) = 5cm

Slant height of cone (l) = r 2 + h2

= (5)2 + (12)2

= =

= 13cm

Curved surface area of a cap =  rl

= 3.14513 = 204.1cm2

Curved surface area of a cap =  rl

= 3.14513 = 204.1cm2

Curved surface area of 15 caps =15 204.1 = 3061.5cm2

Module- 3C A (2023-2024)
 Area of a sheet of paper used for making caps = 25 40 =1000cm2

82% of sheet is used after cutting = 82% of 1000 cm2

82
= 1000 = 820 cm2
100

3061.5
Number of sheet = = 3.73
820

Hence 4 sheets area needed.

51. Curved surface area of a right circular cylinder is 4.4 sqm . if the radius of
the base of the cylinder is 0.7m find its height.

Ans. Let the height of the circular cylinder be h .

Radius (r) of the base of cylinder = 0.7 m

Curved Surface Area of cylinder = 2 rh

4.4 m2 = 2 rh
 22 
4.4 m2 = 2  0.7 
 hm
 7 
44  22 7 
m =  2   h 
10 7 10
 

44 m = (2 22 h)

44
h=
2 22

h = 1m

Therefore, the height of the cylinder is 1m.

Module- 3C A (2023-2024)
 52. The circumference of the trunk of a tree (cylindrical), is 44dm . Find the
volume of the timber obtained from the trunk if the length of the trunk is 5m.

Ans. Let r be the radius of the cylindrical Trunk

Circumference of the trunk = 44dm

Converting dm into m ,

44dm = 4.4 m

The circumference is C = 2r

4.4 = 23.14 r

4.4
r=
2 3.14

r = 0.7

The height is h = 5 m

The volume of the cylinder is

V = r2h

V = 3.140.72 5

V = 7.693m3

Therefore, the volume of the trunk is 7.693cubic meter.

53. If the areas of three adjacent faces of a cuboids are X ,Y and Z. If its
volume is V , prove that V2 = XYZ

Ans. Areas of three faces of cuboid as $x, y, z$

Module- 3C A (2023-2024)
 So, Let length of cuboid be = l

Breadth of cuboid be = b

Height of cuboid be = h

Let, x = l b

y = b h

z = hl

Else write as

xyz = l 2b2h2 .. (i)

If ' V is volume of cuboid = V = lbh

V 2 = l2b2h2 = xyz from (i)

V 2 = xyz

Hence proved.

54. Find the volume of an iron bar has in the shape of cuboids whose length,
breadth and height measure 25cm. 18cm and 6cm respectively. Find also its
weight in kilograms if 1 cu cm of iron weight 100 grams.

Ans. Length of the bar = 25cm

Breadth of the bar = 18cm

Height of the bar = 6cm

 Volume of the iron bar = l b h cu unit

= (2518 6)cu cm

= 2700cu cm

Module- 3C A (2023-2024)
 Weight of the bar = (2700100)gm

= 270000gm

= 270 kg

55. A rectangular piece of paper is 22cm long and 12cm wide. A cylinder is
formed by rolling the paper along its length. Find the volume of the cylinder.

Ans. It is clear that circumference of the base of the cylinder = length of the paper
Let rcm be the radius of the base of the cylinder and its height as hcm .

2r = 22 and h =12cm

22
2  r = 22
7

7
r = cm
2

Volume of the cylinder =  r2 h

22   7 
2
=   12cu cm
7 2

22 7  7 12
= cu cm
7 2 2

= 462cu cm.

56. If the radius of the base of a right circular cylinder is halved, keeping the
height same, find the ratio of the volume of the reduced cylinder to that of
original cylinder.

Ans. Let the radius of the original cylinder = r units

Module- 3C A (2023-2024)
 Height of the original cylinder = h units

 volume of the cylinder =  r2h cu units → (i)

r
Radius of the reduced cylinder = units
2

Height of the reduced cylinder = h units

 r 2
 volume of the cylinder =    h cu units
2

 r 2h
= cu units → (2)
4

From (i) and (ii) we get

volume of cylinder (reduced) r2 h

= 2
volume of the original cylinder r h

1
=
4

Thus, there required ratio = 1: 4

57. A rectangle tank measuring 5m by 4.5m by 2.1m is dug in the centre of a

field 25m by 13.5 m. The earth dug out is spread evenly over the remaining
portion of the field. How much is the level of the field raised?

Ans. Volume of the tank = 5 4.5 2.1cum

= 47.25cm

 Volume of the earth dug = 47.25cum

Area of the field = 2513.5

= 337.5sqm

Module- 3C A (2023-2024)
  Remaining area of the field = (337.5 − 22.5)

= 315sq m

volume of the earth dug out

 Level of the field raised =
remaining area of the field

47.25 4725
= m= cm
315 315

= 15cm

58. A village having a population of 4000 requires 150 litres of water per head
per day. It has a water tank measuring 20 m15m6 m which is full of water.
For how many days will the water tank last?

Ans. Number of days water will last

Volume of tank
=
Total water required per day

Here, l = 20 m,b =15m and h = 6 m

Volume of the tank = l bh

= (20 15 6)m3

=1800 m3

Water required per person per day = 150 litres

Water required for 4000 person per day = (4000150) litres

 1  3
= 4000150 m
 1000 
 

Module- 3C A (2023-2024)
 = 600 m3

Volume of tank
Number of days water will last =
Total water required per day

 1800 m3 
=  600 m3 
 

=3

Thus, the water will last for 3 days.

59. Find the curved surface area of a right circular cone whose slant height is
10cm and base radius is 7cm

Ans. Curved surface area =  rl

22
=  7 10 cm2
7

= 220cm2

60. Find (i) the curved surface area

Ans. The curved  rl surface area of hemisphere of radius 21cm would be = 2r2
22
= 2  21 21cm2
7

= 2772cm2

(ii) Total surface area of a hemisphere of radius 21cm

Ans. The total surface area of the hemisphere = 3 r2

22
= 3  21 21cm2
7

Module- 3C A (2023-2024)
 = 4158cm2

61. The circumference of the base of a cylindrical vessel is 132cm and its
height is 25 cm How many litres of water can it hold? 1000 cm3 = 1l 

Ans. Given circumference of base of cylindrical vessel =132cm

2r =132cm
3
r = 132 = 66  7 = 21cm
2 22

Number of liters of water =  r 2 h

22
=  21 21 25 cm3
7

= 223 21 25cm3

= 34650cm3
 1 
= 34650 litres
 1000 
 

= 34.65 litres

Vessel can hold 34.65 litres.

62. A cubical box has each edge 10cm and another cuboidal box is 12.5cm long,
10cm wide and 8cm high. Which box has the greater lateral surface area and
by how much?

Ans. Side of cubical box = 10 cm

Lateral surface area of cube = 4a2

Module- 3C A (2023-2024)
 4102 = 400cm2

Length of cuboidal box =12.5cm .

Breadth = 10 cm

Height = 8cm

Lateral surface area = 2[l + b]h

= 2[12.5 +10]8

=16 22.5 = 360cm2

Difference = 400 − 360 = 40cm2

Lateral surface area of cuboidal box is greater by 40 cm2

63. A hemi spherical bowl has a radius of 3.5cm . What would be the volume of
water it would contain?
2
Ans. The volume of water the bowl contain =  r
3

Radius of hemisphere = r = 3.5cm

2
The volume of water the bowl can contain =  r
3

2 22
=   3.5 3.5 3.5 cm3
3 7

= 89.8cm3

64. A conical pit of top diameter 3.5m is 12 m deep. What is its capacity in
kiloliters

Module- 3C A (2023-2024)
 Ans. Diameter of conical Pit = 3.5 m

Height of conical pit = h =12 m

Diameter
Radius of conical pit = r =
2

3.5
= m = 1.75 m
2

Capacity of pit =Volume of cone

1
=  r 2h
3

 1 22
= 3  7 1.751.75 12 m 3

 

= 38.5m3

= 38.5 kiloliters

Capacity of pit = 38.5 kiloliters.

65. The diagonals of a cube is 30cm , find its volume

Ans. Let side of cube be a cm

Diagonal = 3a

3a = 30

30
a=
3

3
 30 
Volume of cube = a =  
3

 3

Module- 3C A (2023-2024)
 27000 9000 3
= = cm
3 3 3

66. A cylindrical tank has a capacity of 6160 m3 find its depth if the diameter of
the base is 28m

Ans. Diameter of the base = 28m

28
Radius r = = 14m
2

Volume =  r 2 h = 6160

22
1414 h = 6160
7

6160 7
h= = 10m
221414

Hence depth of tank = 10 m

67. Find the volume of a sphere whose surface area is 154 cm2

Ans. Given surface area of sphere =154cm2

Let radius of the sphere = r cm

22
4  r 2 = 1544  r2
7

= 154r2

154 7 2
= r
4  22

=12.25r

Module- 3C A (2023-2024)
 = 3.5cm

4
Volume of sphere =  r
3

 4 22
= 3  7  3.5 3.5 3.5cm 3

 

=179.67 cm3

68. If the volume of a right circular cone of height 9cm is 48 cm3 Find the
diameter of its base

Ans. Given volume of cone = 48cm3 and height = 9cm

Volume of cone = 48cm3

1
 r 2h = 48
3

1
 r 2 (9) = 48
3

 r 2 3 = 48

48
r2 =
3

r2 = 16

r=

r = (4)2

r = 4 cm

Diameter = 2 Radius
Module- 3C A (2023-2024)
 = 2 4 = 8cm

Thus, the diameter of the base of cone is 8cm .

69. The volume of a cylinder is 69300cm3 and its height is 50cm. Find its curved
surface area

Ans. Volume =  r 2 h = 69300 and h = 50cm

22
  r 2  50 = 69300
7

69300 7
r2 = = 441
22 50

r= = 21cm

 Curved surface area = 2 rh

22
= 2  21 50 = 6600 cm2
7

70. The volume of a cube is 1000 cm3 , Find its total surface area.

Ans. Volume = a3 =1000cm3

a =10cm

Total surface area = 6a2 = 6100

= 600cm2.

Short Answer Questions 3 Mark

Module- 3C A (2023-2024)
 1. A small indoor green house (herbarium) is made entirely of glass panes
(including base) held together with tape. It is 30cm long, 25cm wide and 25cm
high.

(i) What is the surface area of the glass?

Ans. Length (l) of green house = 30 cm

Breadth (b) of green house = 25cm

Height (h) of green house = 25cm

The green house is cuboid and Glass is on the all 6 sides of cuboid greenhouse

Area of glass = Surface area of green house

= 2[lb + lh + bh]

=[2(30 25 + 30 25 + 25 25)]cm2

=[2(750 + 750 + 625)]cm2

= (2  2125)cm2

= 4250cm2

Hence 4250cm2 of the glass is required to make a herbarium.

(ii) How much of tape is needed for all the 12 edges?

Ans. Tape is used at 12 edges.

 Tape is used at 4 lengths, 4 breadths and 4 heights.

 Total length of the tape = 4(l + b + h)

= 2(30 + 25 + 25)

= 320cm

Module- 3C A (2023-2024)
 Hence 320 cm of the tape if needed to fix 12 edges of herbarium.

2. A metal pipe is 77 cm long. The inner diameter of a cross section is 4cm , the
outer diameter being 4.4cm . [See fig.]. Find its:

(i) Inner curved surface area

Ans.

Inner diameter of cross-section = 4cm

Inner diameter
Inner radius of cylindrical pipe = r1 =
2

4
= cm = 2 cm
2
 

Height (h) of cylindrical pipe = 77 cm

Curved Surface Area of inner surface of pipe = 2 r1h

 22 
= 2  2 77 cm2
 7 
 

= 968cm2

Module- 3C A (2023-2024)
 Inner curved surface area is 968cm2

(ii) Outer curved surface area

Ans. Outer diameter of pipe = 4.4 cm

Outer diameter
Outer radius of cylindrical pipe = r2 =
2
 4.4 
= cm = 2.2 cm
 2 
 

Height of cylinder = h = 77 cm

Curved Surface Area of outer surface of pipe = 2 r2h

 22  2
= 2  2.2
 77 cm
 7 

= (2 22 2.211)cm2

=1064.8cm2

Outer curved surface area is 1064.8cm2

(iii) Total surface area

Ans. r1 = 2 cm

r2 = 2.2cm

h = 77 cm

Total surface area = Curved Surface Area of inner cylinder + Curved Surface Area
of outer cylinder + 2 Area of base

Area of base = Area of circle with radius 2.2cm − Area of circle with radius 2cm

=  r22 −  r12

Module- 3C A (2023-2024)
 ((2.2)2 − (2)2 )
22
=
7

22
= (4.84 − 4)
7

22
= (0.84)
7

= 2.64cm2

Total surface area = Curved Surface Area of inner cylinder + Curved Surface Area
of outer cylinder + 2 Area of base

= 968 +1064.8 + 2 2.64

= 2032.8 + 5.28

= 2038.08cm2

Therefore, the total surface area of the cylindrical pipe is 2038.08cm2

3. Curved surface area of a cone is 308cm2 and its slant height is 14cm . Find

(i) radius of the base

Ans. Slant height of cone (l) =14cm

Curved surface area of cone = 308cm2

 rl = 308

22
 r 14 = 308
7

22 r  2 = 308

r =
308 
cm
 2 22 
 

Module- 3C A (2023-2024)
 r = 7 cm

Therefore, the radius of the circular end of the cone is 7 cm .

(ii) total surface area of the cone.

Ans. Total surface area of the cone = Curved surface area + Area of circular
base

= 308 + r2

22
= 308 + (7)2
7

= 462cm2

Therefore, the total surface area of the cone is 462 cm2 .

4. A conical tent is 10 m high and the radius of its base is 24 m . Find:

(i) slant height of the tent.

Ans. Height of the conical tent (h) =10 m

Radius of the conical tent (r) = 24 m

Let the slant height of the tent be l

Slant height of the tent l2 = h2 + r2

l2 = (10)2 + (24)2

l2 = 100 + 576

l2 = 676

l=

l = 262

Module- 3C A (2023-2024)
 l = 26 m

Therefore, the slant height of the tent is 26 m .

(ii) cost of the canvas required to make the tent, if the cost of a m2 canvas is
Rs. 70 .

Ans. Here the tent does not cover the base, So, find curved surface area of tent
Curved surface area of tent =  rl

Here, r = 24m, l = 26m

Curved surface area of tent =  rl

 22 
=  24 26 m2
 7 
 

13728 2
= m
7

Cost of 1m2 canvas = Rs 70

13728
Cost of m2 canvas = Rs 13728  70 
7  7 
 

= Rs137280

Therefore, the cost of the canvas required to make the tent is Rs 137280 .

5. What length of tarpaulin 3m wide will be required to make conical tent of

height 8m and base radius 6 m ? Assume that the extra length of material that
will be required for stitching margins and wastage in cutting is approximately
20cm . (Use  = 3.14)

Ans. Height of the conical tent (h) = 8 m and Radius of the conical tent (r) = 6 m

Module- 3C A (2023-2024)
 Slant height of the tent (l) = r 2 + h2

= (6)2 + (8)2

= 36 + 64

= = 10 m

Area of tarpaulin =Curved surface area of tent =  rl

= 3.14610 =188.4 m2

Width of tarpaulin = 3m

Let Length of tarpaulin = L

 Area of tarpaulin = Length  Breadth

= L3 = 3L

Now, According to question, 3L =188.4

1884.4
L= = 62.8 m
3

The extra length of the material required for stitching margins and cutting is
20cm = 0.2 m.

So, the total length of tarpaulin bought is (62.8 + 0.2)m = 63m

6. A bus stop is barricaded from the remaining part of the road, by using 50
hollow cones made of recycled cardboard. Each cone has a base diameter of
40cm and height 1 m. If the outer side of each of the cones is to be painted and
the cost of painting is Rs. 12 per m2 , what will be the cost of painting all these
cones? (Use  = 3.14 and take 1.04 = 1.02)

Ans. Curved surface area of cone will be painted =  rl

Module- 3C A (2023-2024)
 40
h = 1m; radius = = 20 cm = 0.2 m
2

and let 1 be the slant height,

12 = h2 + r2 =12 + 0.22

1 = = 1.04 =1.02 m

 Curved surface area of 1 cone =  rl

= (3.140.21.02)m2 = 0.64046 m2

 Curved surface area of 50 cones = 500.64046 = 32.028m2

Cost of painting 1m2 = Rs. 12

 Cost of painting 32.028m2 = (1232.028)

= 384.336 m2  384.34

 Cost of painting 50 cones is Rs. 384.84 .

7. The radius of a spherical balloon increases from 7 cm to 14cm as air is being

pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans. I case: Radius of balloon (r) = 7 cm

Surface area of balloon = 4r2 = 4 77 cm2 .. (i)

II case: Radius of balloon (R) = 14cm

Surface area of balloon = 4 R2 = 4 1414cm2 .. (ii)

Now, Ratio [from eq. (i) and (ii)],

CSA in first case 4  7  7 1

= =
CSA in second case 4 14 14 4

Module- 3C A (2023-2024)
 Hence, required ratio = 1: 4

8. A village having a population of 4000 requires 150 litres of water per head
per day. It has a tank measuring 20 m by 15m by 6 m . For how many days will
the water of this tank last?

Ans. Capacity of cuboidal tank = l b h = 20m15m6m

=1800 m3

=18001000 liters

=1800000 liters

Water required by her head per day = 150 liters

Water required by 4000 persons per day =150 4000 = 600000 liters

Capacity of tank (in liter)

Number of days the water will last =
Total water required per day (in liters)

1800000
= =3
600000

Hence water of the given tank will last for 3 days.

9. A godown measures 40 m 25m15m . Find the maximum number of wooden

crates each measuring 1.5m1.25m0.5m that can be stored in the godown.

Ans. Capacity of cuboidal godown = 40m 25m15m =15000m3

Capacity of wooden crate =1.5m1.25m0.5m = 0.9375m3

Volume of godown
Maximum number of crates that can be stored in the godown =
Volume of one crate

Module- 3C A (2023-2024)
 15000
= = 16000
0.9375

Hence maximum 16000 crates can be stored in the godown.

10. Find the minimum number of bricks each measuring 22.5cm11.5cm7.5cm

required to construct a wall 10 m long, 6 m high and 1.5m thick.

Ans. Volume of one cuboidal brick = l b h

= 22.5cm11.5cm7.5cm3

= 1940.625cm3

= 0.001940625m3

Volume of cuboidal wall =10 m6 m1.5m

= 90 m3

Volume of wall
Minimum number of bricks required =
Volume of a brick

90
=
0.001940625

90
=
1940625
1000000000

90000000000
= = 46376.81
1940625

= 46377 [Since bricks cannot be in fraction]

11. The circumference of the base of a cylindrical vessel is 132cm and its
height is 25cm How many litres of water can it hold?

Module- 3C A (2023-2024)
 Ans. Height of vessel = (h) = 25cm

Circumference of base of vessel = 132cm

 2r =132

22
 2  r = 132
7

132 7
r= = 21cm
2 22

Now, Volume of cylindrical vessel =  r 2 h

22
=  21 21 35 = 34650 cm3
7

34650
= liters
1000

= 34.65 liters

12. The inner diameter of a cylindrical wooden pipe is 24cm and its out
diameter is 28 cm . The length of the pipe is 35cm . Find the mass of the pipe, if
1cm3 of wood has a mass of 0.5g

Ans. Inner diameter of pipe = 28cm

24
 Inner radius of pipe (r) = = 12 cm
2

And Outer diameter of pipe = 28cm

28
 Outer radius of pipe (R) = = 14 m
2

Length of pipe (h) = 35cm

Volume of wood = Volume of outer cylinder −Volume of inner cylinder

Module- 3C A (2023-2024)
 =  R2h − r 2h = h (R2 − r 2 )

22
=  35 (14)2 − (12)2 
7

=110[196 −144] =11052 = 5720cm3

Weight of 1cm3 of wood = 0.6 g

 Weight of 5720cm3 of wood = 0.65720

= 3432g = 3.432 kg

Therefore, mass of pipe is 3.432 kg

13. A soft drink is available in two packs (i) a tin can with a rectangular base
of length 5cm and width 4cm , having height of 15cm

Ans. Given, Length = 5cm

Width = 4cm

Height = 15cm

Volume of the tin can V = lbh

= 5 415 = 300cm3

(ii) a plastic cylinder with circular base of diameter 7 cm and height 10cm .
Which container has greater capacity and how much?
7
Ans. Given, Diameter = 7 cm , Height = 10 cm  = Volume =  r2 h
2

22 7 7
=  
7 2 2

Difference = 385cm3 − 300cm3 = 85cm3

Module- 3C A (2023-2024)
 Hence, Cylinder container has greater capacity by 85 cubic cm.

14. It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel
10 m deep. If the cost of painting is at the rate of Rs. 20 per m2 , find:

(i) inner curved surface area of the vessel.

Ans. Total cost to paint inner curved surface area of the vessel = Rs. 2200

Rate = Rs. 20 per square meter

Total cost
Inner curved surface area of vessel =
Rate

2200
= = 110 m2
20

(ii) radius of the base.

Ans. Depth of the vessel (h) =10 m

Now, Inner surface area of vessel = 110 m2

 2 rh =110

22
 2  r 10 = 110
7

110 7
r= = 1.75 m
2 2210

(iii) capacity of the vessel.

Ans. Since r =1.75m and h = 10 m

 Capacity of vessel = Volume of cylinder =  r 2 h

22
= 1.751.7510 = 96.25 m3
7

Module- 3C A (2023-2024)
 = 96.25kl

15. The capacity of a closed cylindrical vessel of height 1m is 15.4 litres. How
many square meters of metal sheet would be needed to make it?

Ans. Height of the vessel (h) = 1m

Capacity of vessel = 15.4 liters

15.4
= kilo liters
1000

= 0.0154m3

  r 2 h = 0.0154

22
  r 2 1 = 0.0154
7

0.0154 7
 r2 =
22

 r2 = 0.00077 = 0.0048

 r = 0.07 m

Now, Area of metalsheet required= TSA of cylindrical vessel

= 2r(r + h)

22
= 2  0.07(1+ 0.07)
7

44
=  0.07 1.07
7

= 0.4708m2

Module- 3C A (2023-2024)
 16. Find the capacity of a conical vessel with:

(i) Radius 7 cm , Slant height 25cm

Ans. Given: r = 7 cm, l = 25cm

h = l2 − r2

= (25)2 − (7)2

= 625 − 49

= = 24 cm

1
Capacity of conical vessel =  r h
2

1 22
=   7  7  24 = 1232 cm3
3 7

= 1.232 liters

(ii) Height 12cm , Slant height 13cm

Ans. Given: h =12cm, l =13cm

r = l 2 − h2 = (13)2 − (12)2

= 169 −144

= = 5cm

1
Capacity of conical vessel =  r h
2

1 22 2200 3
=   5 512 = cm
3 7 7

2200 1 liters
= 
7 1000

Module- 3C A (2023-2024)
 11
= liter
35

17. If the triangle ABC in question 7 above is revolved about the side 5cm ,
then find the volume of the solid so obtained. Find, also, the ratio of the
volume of the two solids obtained.

Ans. When right angled triangle ABC is revolved about side 5cm , then the solid
formed is a cone.

In that cone, Height (h) = 5cm

And radius (r) =12cm

1
Therefore, Volume of cone =  r h
2

1
=  1212 5
3

= 240cm3

Now, Volume of cone in Q. No. 7 = 100 = 5

Volume of vone in Q. No. 8 240 12

 Required ratio = 5 :12

18. The diameter of the moon is approximately one-fourth the diameter of the
earth. What fraction is the volume of the moon of the volume of the earth?

Ans. Let diameter of earth be x

x
 Radius of earth (r) =
2

Module- 3C A (2023-2024)
 4
Now, Volume of earth =  r
3

4 x x x 1 4
=     =   x3
3 2 2 2 8 3

According to question,
1
Diameter of moon =  Diameter of earth
4

1 x
= x =
4 4

x
Radius of moon (R) =
8

4
Now, Volume of Moon =  R
3

3
4 x x x 1 4 3
=     =  x
3 8 8 8 512 3
1 1 4 3 
=   x
64 8 3 

1
=  Volume of Earth
64

1
Volume of moon is th the volume of earth.
64

19. How many litres of milk can a hemispherical bowl of diameter 10.5 hold?

Ans. Diameter of hemispherical bowl = 10.5cm

10.5
 Radius of hemispherical bowl (r) = = 5.25 cm
2

Module- 3C A (2023-2024)
 2
Volume of milk in hemispherical bowl =  r
3

2 22
=   5.25 5.25 5.25
3 7

2 22 525 525 525

=    
3 7 100 100 100

21 21
= 11  = 303.187 cm3
4 4

303.187
= liters
1000

= 0.303187 liters = 0.303 liters

20. Find the volume of a sphere whose surface area is 154 cm2 .

Ans. Surface area of sphere =154cm2

 4r2 =154

22
 4  r 2 = 154
7

154 7 49
 r2 = =
4  22 4

7
 r = cm
2

4 4 22 7 7 7
Now, Volume of sphere =  r =    
3

3 3 7 2 2 2

1 539
= 11 49 =
3 3

= 179 2 cm3
3

Module- 3C A (2023-2024)
 21. A wooden bookshelf has external dimensions as follows: Height =110cm ,
Depth = 25 cm, Breadth = 85cm . The thickness of the planks is 5cm
everywhere. The external faces are to be polished and the inner faces are to be
painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is
10 paise per cm2 , find the total expenses required for polishing and painting
the surface of the bookshelf

Ans. External faces to be polished

=Area of six faces of cuboidal bookshelf − 3 (Area of open portion ABCD)

= 2(110 25 + 2585 + 85110) − 3(7530)

= 2(2750 + 2125 + 9350) − 3 2250

= 214225 − 6750

= 28450 − 6750

Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise.

= Rs. 0.20 per cm2

= Rs. 0.20 21700 = Rs. 4340

Here, three equal five sides inner faces.

Therefore, total surface area = 3[2(30 + 75)20 + 3075] [ Depth = 25 − 5 = 20cm]

= 3[2105 20 + 2250] = 3[4200 + 2250]

= 36450 =19350cm2

Now, cost of painting inner faces at the rate of 10 paise i.e. Rs. $0.10$ per cm2 .

= Rs 0.1019350 = Rs. 1935

Module- 3C A (2023-2024)
 22. If diameter of a sphere is decreased by 25% then what percent does its
curved surface area decrease?

Ans. Diameter of original sphere = D = 2R

D
R=
2

Curved surface area of original sphere = 4 R2

 D 2
= 4   =  D2
2

25
According to the question, Decreased diameter = 25% of D = D
100

D
=
4

D 3D
 Diameter of new sphere = D − =
4 4

3D
 Radius of new sphere =
8

 3D 2
Now, curved surface area of new sphere = 4  r = 4  2

 8 

9 2
= D
16

9
Change in curved surface area =  D −
2
D2
16

7
=  D2
16

Module- 3C A (2023-2024)
 Change in curved surface area
Percent change in the curved surface area =
Curved surface area of origianal sphere
7
100 = 7  D2 100
D 2

16

7
= 100 = 43.75%
16

23. The surface area of cuboids is 3328 m2 ; its dimensions are in the ratio
4 : 3: 2.Find the volume of the cuboid.

Ans. Let the dimensions of the cuboid be 4x,3x and 2x meters

Surface area of the cuboid = 2(4x 3x + 3x  2x + 2x  4x) sq m

= 2 (12x2 + 6x2 + 8x2 ) sq m

= 52x2sqm → (i)

Given surface area = 3328 sq m

From (i) and (ii) we get

52x2 = 3328

3328
or x2 = = 64
52

or x = 8

4x = 32,3x = 24 and 2x =16

Thus, the dimensions of the cuboid are 32 m, 24 m and 16 m

 Volume of the cuboid = (32 2416)m3

=12288cu m

Module- 3C A (2023-2024)
 24. The volume of a rectangular slower of stone is 10368dm3 and is dimensions
are in the ratio of 3:2:1. (i) Find the dimensions (ii) Find the cost of polishing
its entire surface @ Rs. 2 per dm2 .

Ans. Let the length of the block be 3x dm

Width = 2 dm and height = xdm

Volume of the block =10368dm3

3x 2x x =10368

or x3 = 10368
6

= 1728

 x = 1728

= 121212 =12

also 2x = 24 and 3x = 36

Thus, dimensions of the block are 36 dm, 24 dm and 12 dm

Surface area of the block = 2(36 24 + 2412 + 3612) dm2

= 2(864 + 288 + 432)dm2

= 21584dm2

= 3168dm2

Cost of polishing the surface = Rs(23168)

= Rs. 6336

Module- 3C A (2023-2024)
 25. In a cylindrical drum of radius 4.2m and height 3.5m , how many full bags
of wheat can be emptied if the space required for each bag is 2.1cum .
42
Ans. Radius of the drum = 4.2 m = m
10

35
Height of the drum = 3.5 m = m
10

 Volume of the drum = r2h cu units

 22 42 42 35 
=      cu m ..(i)
7 10 10 10
 

21
Volume of wheat in each bags = 2.1cu m = cu m
10

volume of drum
=
volume of wheat in each bag

22 42 42 35
  
= 7 10 10 10
21
10

924
= = 92.4
10

= 92

Hence the number of full bags is 92 .

26. The inner diameter of a cylindrical wooden tripe is 24cm. and its outer
diameter is 28cm . the length of wooden tripe is 35cm . find the mass of the
tripe, if 1cu cm of wood has a mass of 0.6g .

Ans. Inside diameter of the pipe = 24 cm

Outside diameter of the pipe = 28cm

Module- 3C A (2023-2024)
 Length of the pipe = 35cm = h
28
Outside radius of the pipe = cm = 14 cm = R
2
Volume of the wood = Externalvolume − Internal volume

=  r 2 h − 2l

=   35(142 −122 ) cu cm

22
=  35(14 +12)(14 −12) cu cm
7

= 5720 cu cm

Mass of 1 cu cm = 0.6g

 Mass of the pipe = (0.65720)g

= 3432 g

= 3.432 kg

27. A patient in a hospital is given soup daily in a cylindrical bowl of a

diameter 7 cm . If the bowl is filled with soup to height of 4cm. How much soup
the hospital has to prepare daily to serve 250 patients?

Ans. Diameter of the bowl = 7 cm.

7
Radius of the bowl = cm
2

Height up to which soup is filled (h) = 4cm.

Volume of the soup in one bowl =  r 2 h

22 7 7
=    4 cu cm
7 2 2

Module- 3C A (2023-2024)
 =154 cu cm

 soup given to one patient =154 cu cm .

Soup given to 250 patients = 250154 cu cm

= 38500 cu cm

38500
= ltrs
1000

= 38.5 ltrs.

Hence the hospital has to prepare 38.5 litre daily to serve 250 patients.

28. The diameter of a roller is 84cm and its length is 120 cm. It takes 500
complete revolutions to move once over to level a playground.

(a) Find the area of playground in sq m .

84
Ans. R = Radius of the roller =
2

Area = 42cm= 0.42m

H = length of the roller =120cm. =1.2 m .

Area covered in the revolution = 2rh sq unit

2 22  0.42 1.2
=
7

= 3.168 sqm

 Area covered in 500 revolutions = 5003.168 sq m

=1584 sqm

Thus, area of playground =1584sqm .

Module- 3C A (2023-2024)
 (b) Determine the cost of leveling the playground at the rate of Rs 1.75 per sq
m.

Ans. cost of leveling 1 sq m . of playground = Rs 1.75

Cost of total leveling = Rs(15841.75)

= Rs 2772

29. A metal cube of edge 12cm is melted and formed into three similar cubes.
If the edge of two smaller cubes is 6cm and 8cm , find the edge of the third
smaller cube (Assuming that there is no loss of metal during melting).

Ans. Volume of cube with edge 12cm = (12)3 cu cm .

=1728 cu cm ..........(1)

Volume of the first smaller cube with edge 6cm = (6)3cu cm

= 216 cu cm ........ (2)

Volume of the second smaller cube with edge 8 cm. = (8)3 cu cm

= 512cu cm....(3)

Let the edge of the third smaller cube be a cm .

 Volume of the third smaller cube == 92 cm3..............(4)

216 + 512 + a3 =1728 [using (1) and (2)]

By the given condition.

area 728a3 = 1728

Area a3 =1728 − 728 =1000 = (10)3

a = 10

Module- 3C A (2023-2024)
 Thus, the edge of the third required cube is 10cm.

30. How many bricks, each measuring 18cm by 12cm by 10cm will be required
1
to build a wall 15m long 6dm wide and 6.5m high when of its volumes
10
occupied by mastar? Please find the cost of the bricks to the nearest rupees, at
Rs 1100 per 1000 bricks.

Ans. Length of the wall =15m. =1500cm .

Width of the wall = 6 dm. = 60cm.

Height of the wall = 6.5m. = 650cm.

 Volume of the wall = (150060650) cucm

= 58500000 cu cm. → (I)


Volume occupied by master =  10  58500000 cu cm
1
 

= 5850000 cu cm. → (ii)

 Volume occupied by bricks = (i) −(ii)

= (58500000 − 5850000) cucm

= 52650000 cu cm. → (iii)

Volume of a brick = (181210) cu cm

= 2160 cu cm. → (iv)

 No of brick required

52650000
=
2160

Module- 3C A (2023-2024)
 = 24375

cost of 1000 bricks = Rs 1100

243751100
Total cost = Rs
1000

= Rs 26812.50

31. A river 3m deep and 40 m wide is flowing at the rate of 2 km per hour.
How much will fall into the sea in a minute?

Ans. Depth of river = 3m

Water of the river = 40 m

Rate of flow of water = 2 km / hr = 2000 m / hr

 Volume of water flowing in one hour

= 2000 403

= 240000m3

240000
Hence, Volume of water flowing in one minute = = 4000 m3
60

32. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5cm . then find

(i) radius of its base

Ans. Given lateral surface of cylinder = 94.2cm2

2 rh = 94.2cm2

H = 5cm

Module- 3C A (2023-2024)
 2r 5 = 94.2

94.2 94.2
r= = cm
10 10 3.14

R = 3cm

(ii) its volume [ = 3.14]

Ans. Volume of cylinder =  r2 h

= 3.1432 5

=141.3cm3

33. A shot put is a metallic sphere of radius 4.9cm If the density of the metal is
7.8g per cm3 Find the mass of the shot put.

4
Ans. Volume of sphere = r and radius r = 4.9cm
3

4 22
=   4.9 4.9 4.9 cm3
3 7

= 493cm3

Mass of 1cm3 of metal is 7.8g

Mass of the shot put = volume  density

= 7.8 493g

= 3845.44g = 3.85kg

34. The capacity of a hemispherical tank is 155.232l. Find its radius.

Module- 3C A (2023-2024)
 2
Ans. Capacity of tank = Its Volume =  r
3

2
 r 3 = 155.232l
3

=155.2321000cm3

=155232cm3

2 22 3
  r = 155232 ,
3 7

155232  3 7
r3 =
2 22

r3 = 352837

r3 = (23 7)3

r = 237 = 42cm

Hence radius of tank = 42 cm

35. What length of tarpaulin 3m wide will required to make conical tent of
height 8m and base radius 6 m ? Assume that the extra length of material that
will be required for stitching margins and wastage in cutting is approximately
20cm

Ans. Here h = 8 m and r = 6 m

l = r 2 + h2 = 36 + 64 = 10 m

Curved surface area =  rl

= 3.14610 =188.4 m2

Length of tarpaulin required = area = 188.4

width 3

Module- 3C A (2023-2024)
 = 62.8m

Extra length required for wastage = 20cm = 0.2 m

Hence, total length required = 62.8 + 0.2

= 63m

36. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm How

much medicine (inmm3 ) is needed to fill this capsule?

3.5
Ans. Given radius of capsule = mm
2

4 3
Amount of medicine = Volume of capsule =  r
3

4 22 (3.5)3
=   3
mm
3 7 2

4 22 3.5 3.5 3.5

=    
3 7 2 2 2

= 22.46 mm3( approx )

37. A wall of length 10 m was to be built across an open ground. The height of
wall is 4m and thickness of the wall is 34cm . If this wall is to be built up with
bricks whose dimensions are 24cm12cm8cm. How many bricks would be
required

Ans. Length of wall =10 m =1000cm

Thickness = 24 cm

Height = 4 m = 400cm

Module- 3C A (2023-2024)
 Volume of wall = length  thickness  height =1000 24 400cm3

Now each brick is a cuboid with length = 24 cm

Breadth = 12 cm and height = 8cm

Volume of each brick =1b h = 24128cm3

volume of the wall

Number of bricks required =
volume of each brick

1000  24  400
= = 4166.6
2412 8

The wall requires 4167 bricks.

38. The pillars of a temple are cylindrically shaped if each pillar has a circular
base of radius 20cm and height 10 m . How much concrete mixture would be
required to build 14 such pillars?

Ans. Radius of base of cylinder = 20 cm

Height of pillar =10 m =1000cm

Volume of each cylinder =  r 2 h

22
=  20 201000 cm3
7

8800000 3
= cm
7

8.8 3
= m
7

 Volume of 14 pillars = volume of each cylinder 14

8.8
= 14 cm3 = 17.6 m3
7

Module- 3C A (2023-2024)
 So, 14 pillars would need 17.6m3 of concrete mixture.

39. A right triangle ABC with sides 5cm,12cm , and 13cm is revolved about the
side12 cm , find the volume of the solid so obtained

Ans. The solid obtained by revolving the given right triangle is a right circular
cone with radius = 5cm

And height = 12 cm
1 2
 Volume of solid =  r h
3

1
=   52 12 = 100 cm3
3

40. The inner diameter of a circular well is 3.5cm . It is 10 m deep find.

(i) Its inner curved surface area.

Ans. Given Inner diameter of well = 3.5 m

3.5 7
 Inner radius = = m
2 4

7
r= m and depth h = 10m
4

(i)  Inner surface area = 2 rh

 22 
= 2 1.7510 m 2
 7 
 

=  2 7  100  10 m 2
22 175

 

Module- 3C A (2023-2024)
  25  2
= 2 22 m
 10 
 

= 110 m2

(ii) the cost of plastering this curved surface at the rate of Rs 40 per

Ans. The cost of plastering is Rs 40 per m2

 Cost of plastering this surface area = Rs 40110

= Rs 4400

41. A Godown measures 40m 25m10m . Find the maximum number of

wooden crates each measuring 10m1.25m0.5m that can be stored in the go
down

Ans. Dimensions of Godown

= 40m 25m10m

Volume of Godown = 40m 25m10m =10000m3

volume of wooden carts =10m1.25m0.5m = 6.25m3

10, 000
No. of wooden crates =
6.25

= 800

Hence, 800 wooden crates are required.

42. The volume of a right circular cylinder is 576cm3 and radius of its base is
8cm . Find the total surface area of the cylinder.

Ans. Volume of cylinder = 576cm3

Module- 3C A (2023-2024)
 r = 8cm

Volume of cylinder =  r 2 h

 r 2 h = 576

576 576
h= = 2 =9
r2 8

H = 9cm

 Total surface area = 2r(r + h)

22
= 2 (8 + 9)cm2
7

16 2217 2
= cm
7

= 854.989cm

Long Answer Questions 4 Mark

1. 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 25cm by 20 cm by 5cm and the smaller of dimensions 15cm by
12cm by 5cm. 5% of the total surface area is required extra, for all the
overlaps. If the cost of the card board is Rs. 4 for 1000 cm2 , find the cost of
cardboard required for supplying 250 boxes of each kind.

Module- 3C A (2023-2024)
 Ans. Given, Length of bigger cardboard box (L) = 25cm

Breadth ( B ) = 20 cm and Height (H) = 5cm

Total surface area of bigger cardboard box

= 2(LB+ BH + HL)

Substitute values

= 2(25 20 + 205 + 5 25)

= 2(500 +100 +125)

=1450cm2

5% extra surface of total surface area is required for all the overlaps.

5
 5% of 1450 = 1450 = 72.5 cm2
100

Now, total surface area of bigger cardboard box with extra overlaps

=1450 + 72.5 =1522.5cm2

 Total surface area with extra overlaps of 250 such boxes

= 2501522.5 = 380625cm2

Since, Cost of the cardboard for 1000cm2 = Rs. 4

4
Now, Cost of the cardboard for 1cm2 = Rs.
1000

4
Cost of the cardboard for 380625cm2 = Rs.  380625 = Rs. 1522.50
1000

Module- 3C A (2023-2024)
 Now length of the smaller box (l) =15cm ,

Breadth (b) =12cm and Height (h) = 5cm

Total surface area of the smaller cardboard box

= 2(lb + bh + hl)

Substitute values

= 2(1512 +125 + 515)

= 2(180 + 60 + 75)

= 2315 = 630cm2

5% of extra surface of total surface area is required for all the overlaps.

5
Thus, 5% of 630 =  630 = 31.5 cm2
100

Total surface area with extra overlaps = 630 + 31.5 = 661.5cm2

Now Total surface area with extra overlaps of 250 such smaller boxes

= 661.5 250 =165375cm2

Cost of the cardboard for 1000cm2 = Rs. 4

4
Cost of the cardboard for 1cm2 = Rs.
1000

4
Cost of the cardboard for 165375cm2 = Rs. 165375 = Rs. 661.50
1000

Module- 3C A (2023-2024)
 Therefore, Total cost of the cardboard required for supplying 250 boxes of each
kind

=Total cost of bigger boxes + Total cost of smaller boxes

= Rs. 1522.50 + Rs. 661.50

= Rs. 2184

2. Find

(i) the lateral or curved surface area of a petrol storage tank that is 4.2 m in
diameter and 4.5m high.

Ans. Diameter of cylindrical petrol tank = 4.2 m

4.2
Thus, Radius of the cylindrical petrol tank = = 2.1m
2

And Height of the tank = 4.5m

Therefore, Curved surface area of the cylindrical tank

22
= 2 rh = 2  2.11.45 = 59.4 m2
7

1
(ii) how much steel was actually used if of the steel actually used was
12
wasted in making the tank?

Ans. Let the actual area of steel used be x meters

1
Since of the actual steel used was wasted, the area of steel which has gone into
12
the tank.
1 11
x− x= x
12 12

Module- 3C A (2023-2024)
 11
x = 59.4
12

12
 x = 59.4 = 64.8 m2
11

Hence, the steel actually used is 64.8 m2 .

3. A hemispherical bowl made of brass has inner diameter 10.5cm . Find the
cost of tinplating it on the inside at the rate of Rs. 16 per 100 cm2 .

Ans. Inner diameter of bowl =10.5cm

10.5
Thus, Inner radius of bowl (r) = = 5.25 cm
2

Now, Inner surface area of bowl = 2r2

22
= 2  5.25 5.25
7

22 21 21 693 2
= 2   = cm
7 4 4 4

cost of tin-plating per 100cm2 = Rs. 16

16
Then, Cost of tin-plating per 1cm2 =
100
16 693
Therefore, Cost of tin-plating per 693 cm2 =  = Rs. 27.72
4 100 4

4. The diameter of the moon is approximately one fourth the diameter of the
earth. Find the ratio of their surface areas.

Ans. Let diameter of Earth = x

Module- 3C A (2023-2024)
 x
Thus, Radius of Earth (r) =
2

x x
Surface area of Earth = 4  r = 4   =  x
2 2

2 2

1 x
Now, Diameter of Moon = th of diameter of Earth =
4 4

x
Thus, Radius of Moon(r) =
8

x x
Surface area of Moon = 4r2 = 4   =  x
2

8 8 16

Surface area of Moon  x2  x2 1 1

Now, Ratio = = =  2 =
Surface area of Earth 16 16  x 16
 x2

Therefore, required ratio = 1:16

5. A solid cube of side 12cm is cut into eight cubes of equal volume. What will
be the side of the new cube? Also, find the ratio between their surface areas.

Ans. Volume of solid cube = ( side )3 = (12)3 =1728cm3

1
Volume of each new cube = (Volume of original cube)
8

Module- 3C A (2023-2024)
 1
= 1728 = 216 cm3
8

Side of new cube = 3 216 = 6 cm

Now, Surface area of original solid cube = 6( side )2

= 61212 = 864cm2

Now, Surface area of original solid cube = 6 (side) 2

= 666 = 216cm3

Now according to the question,

Surface area of original cube 864 4

= =
Surface area of new cube 216 1

Hence, required ration between surface area of original cube to that of new cube
= 4 :1.

6. The volume of a right circular cone is 9856cm3. If the diameter of the base if
28cm , find:

(i) Height of the cone

Ans. Diameter of cone = 28cm

Radius of cone = 14 cm

Module- 3C A (2023-2024)
 Volume of cone = 9856cm3

1
  r 2 h = 9856
3

1 22
  1414 h = 9856
3 7

9856  3 7
h= = 48 cm
22 14 14

(ii) Slant height of the cone

Ans. Slant height of cone (l) = r 2 + h2

= (14)2 + (48)2 = 196 + 2304

= = 50 cm

Module- 3C A (2023-2024)
 (iii) Curved surface area of the cone.

Ans. Curved surface area of cone =  rl

22
= 14 50 = 2200 cm2
7

7. The front compound wall of a house is decorated by wooden spheres of

diameter 21cm , placed on small supports as shown in figure. Eight such
spheres are used for this purpose and are to be painted silver. Each support is
a cylinder of radius 1.5cm and height 7 cm and is to be painted black. Find he
cost of paint required if silver paint costs 25 paise per cm2 and black paint
costs 5 paise per cm2

Ans. Diameter of a wooden sphere = 21cm.

21
Then, Radius of wooden sphere (R) = cm
2

And Radius of the cylinder (r) =1.5cm

Surface area of silver painted part = Surface area of sphere - Upper part of cylinder
for support

= 4 R2 −  r 2

=  (4R 2 − r 2 )

Module- 3C A (2023-2024)
 Substitute values
22   21   15  
2 2

=   4   −  10  
7 2
     

22  4  441 9 
=  −
7  4 4 
22 1764 − 9 
=
7  4 

22 1755
=  = 1378.928 cm2
7 4

Surface area of such type of 8 spherical part = 81378.928

=11031.424cm2

Since, Cost of silver paint over 1cm2 = Rs. 0.25

Therefore, Cost of silver paint over 11031.928cm2 = 0.2511031.928 = Rs. $2757.85$

Now, curved surface area of a cylindrical support = 2 rh

22 15
= 2   7 = 66 cm2
7 10

Curved surface area of 8 such cylindrical supports = 668 = 528cm2

Since, Cost of black paint over 1cm2 of cylindrical support = Rs. $0.50$

Therefore, Cost of black paint over 528cm2 of cylindrical support = 0.50528

= Rs. 26.40

Therefore, Total cost of paint required = Rs. 2757.85 + Rs. 26.4 = Rs.2784.25

Module- 3C A (2023-2024)
 8. The difference between outside and inside surface of a cylindrical metallic
tripe 14 cm long is 44sqcm . If the tripe is made of 99cucm of metal, find the
outer and inner radius of the tripe.

Ans. Let r1 cm and r2 cm can be the inner and outer radii respectively of the pipe

Area of the outside surface = 2 r2h sq unit

Area of the inside surface = 2 r1h unit

By the given condition

2 r2h − 2 r1h = 44

or 2h (r2 − r1 ) = 44
22
2 14 ( r − r ) = 44( h = 14 cm)
2 1
7

Or, 88(r2 − r1 ) = 44
1
(r − r ) = → (2)
2 1
2

Again volume of the metal used in the pipe ==  ( r22 − r12 )h cu units
22
(r 2 − r 2 )14 = 99
2 1
7

or, 44 ( r 2 − r 2 ) =
99 9
= → (2)
2 1
44 4

Divide (1) by (2)

(r2 − r2 ) = 9  1
2 1

r2 − r1 4 2

Module- 3C A (2023-2024)
 9
Or, r (r2 − r1 )(r2 + r1 ) = 9  2 (r + r ) =
(r −r ) 4 1 2 1
2
2 1

1
Also, ( r − r ) = [ From(1) ]
2 1
2

2r2 = 5

Adding
5
r =
2
2
5 9
And, +r =
1
2 2
9 5
Therefore, r = −
1
2 2

Or, r1 = 2

Thus, outer radius = 2.5cm and inner radius = 2cm .

9. The ratio between the radius of the base and height of a cylinder is 2 : 3.
Find the total surface area of the cylinder if its volume is 1617 cm3

Ans. Let the radius of the base of the cylinder be 2xcm .

Thus, Height of the cylinder = 3x cm .

Volume of the cylinder = r2h cu units

22
= (2x)2  3x cu cm.
7

22
=  4x2  3x cu cm.
7

Module- 3C A (2023-2024)
 264
= x3cucm
7

By the given condition

264
x3 = 1617
7

1617  7 49  7  7 3
x =
3
= = 
264 8 2

7
Thus, radius = 2  = 7 cm
2

7 21
And height = 3 = cm
2 2

Total surface area = 2r(r + h) sq units

22  21 
= 2 7 7 + sq cm.
7  2 
 

35
= 44  sq cm
2

= 770sqcm.

Thus total surface area of the cylinder is 770sqcm.

10. Twenty-seven solid iron spheres, each of radius r and surface area S are
melted to form a sphere with surface area S' find the

(i) radius r ' ' of the new sphere

Ans. Total volume of 27 iron spheres = Volume of new sphere

4
Volume of each original sphere =  r
3

Module- 3C A (2023-2024)
 4 108
Volume of 27 spheres = 27   r = r3
3

3 3

108 3
Volume of new sphere = r
3

4 108
 ( r ) =
3
r3
3 3

( r ) 108 3
3
= r3 
3 4

= 27r3

Therefore, r = 3r .

(ii) ratio of S and S 

Ans. Surface area of original sphere (S) = 4r2

Surface area of new sphere (S  ) = 4 (r )

2

= 4 (3r)2

= 36 r2

 4r 2 1
Therefore, Ratio of S and S = = = 1: 9.
36r2 9

11. 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 and the smaller of dimensions 15cm12cm5cm for all the
overlaps, 5% of the total surface area is required extra. If the cost of
cardboard is Rs 4 for 1000 cm2 . Find the cost of cardboard required for
supplying 250 boxes of each kind.

Ans. Given dimensions of bigger box

Module- 3C A (2023-2024)
 = 25cm 20cm5cm

Total surface area of bigger box

= 2[25 20 + 205 + 255]cm2

= 2[500 +100 +125]cm2 = 2725 =1450cm2

Extra cardboard for packing = 5% of 1450 cm2

5
= 1450 = 72.5 cm2
100

Cardboard used for making box =1450 + 72.5 =1522.5cm2

Dimensions of smaller box =15cm12cm5cm

Total surface area of smaller box = 2[1512 +1215 +155]cm2

= 2[180 + 60 + 75]cm2

= 2315cm2 = 630cm2

Extra cardboard for packing = 5% of 630

5
=  630 = 31.5 cm2
100

Total area of cardboard = 630 + 31.5 = 661.5cm2

Total cardboard used for making 2 boxes

= (1522.5 + 661.5)cm2 = 2184cm2

Cardboard used for making 250 boxes = 250 2184 = 546000cm2

4
Cost of cardboard =  546000 = Rs.2184
1000

Module- 3C A (2023-2024)
 12. A hollow spherical shell is made of a metal of density 9.6g / cm3 . The
external diameter of the shell is 10cm and its internal diameter is 9cm . Find

(i) Volume of the metal contained in the shell

Ans. External diameter of the spherical shell= 10cm

External radius R = 5cm

Internal diameter = 9cm

9 9
Internal radius = cm r = cm
2 2

4
Volume of the metal =  R3 − r3  cm3
3

Substitute values
4  9 
3

=  5 −    cm3
3

3   2  

=  125 −
4 22 729  3
cm
3 7  8 

88 271 3
=  cm = 141.95 cm3
21 8

(ii) Weight of the shell.

Ans. Weight of the shell = Volume  Density

= 141.95cm3 9.6gm / cm3

=1363gm

=1.363kg

(iii) Outer surface area of the shell.

Module- 3C A (2023-2024)
 Ans. Outer surface area = 4r2

= 4 (5)2

22
= 4  25
7

2200
= = 314.389 cm2
7

Module- 3C A (2023-2024)