NCERT Exemplar Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1

Chapter 12 Surface Areas and Volumes NCERT Exemplar Solutions Exercise 12.1 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 12 Surface Areas and Volumes Exercise 12.1
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 12.2 Exercise 12.3 Exercise 12.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 12.1 Solutions

Multiple Choice Questions:

Choose the correct answer from the given four options:

1. A cylindrical pencil sharpened at one edge is the combination of
(A) a cone and a cylinder
(B) frustum of a cone and a cylinder
(C) a hemisphere and a cylinder
(D) two cylinders

Solution

(A) a cone and a cylinder 

The Nib of a sharpened pencil = conical shape
And,
The rest of the part of a sharpened pencil = cylindrical
Therefore, a pencil is a combination of cylinder and a cone.

2. A surahi is the combination of:
(A) a sphere and a cylinder

(B) a hemisphere and a cylinder

(C) two hemispheres

(D) a cylinder and a cone

Solution

(A) a sphere and a cylinder 

The top part of surahi = cylindrical shape

Bottom part of surahi = spherical shape

Therefore, surahi is a combination of Sphere and a cylinder.

3. A plumbline (sahul) is the combination of

(A) a cone and a cylinder

(B) a hemisphere and a cone

(C) frustum of a cone and a cylinder

(D) sphere and cylinder

Solution

(B) a hemisphere and a cone 

The upper part of plumb line = hemispherical,

The bottom part of plumb line = conical

Therefore, it is a combination of hemisphere and cone.

4. The shape of a glass (tumbler) is usually in the form of 

(A) a cone

(B) frustum of a cone

(C) a cylinder

(D) a sphere

Solution

(B) frustum of a cone

The shape of glass is a frustum or specifically, an inverted frustum. 

5. The shape of a gilli, in the gilli-danda game, is a combination of

(A) two cylinders

(B) a cone and a cylinder

(C) two cones and a cylinder

(D) two cylinders and a cone

Solution

(C) two cones and a cylinder

The left and right part of a gilli = conical

The central part of a gilli = cylindrical

Therefore, it is a combination of a cylinder and two cones.

6. A shuttle cock used for playing badminton has the shape of the

combination of

(A) a cylinder and a sphere

(B) a cylinder and a hemisphere

(C) a sphere and a cone

(D) frustum of a cone and a hemisphere

Solution 
(D) frustum of a cone and a hemisphere

The cork of a shuttle = hemispherical shapes

The upper part of a shuttle = shape of frustum of a cone.

Therefore, it is a combination of frustum of a cone and a hemisphere.

7. A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Solution 

(A) a frustum of a cone 

When a cone is divided into two parts by  a plane through any point on its axis parallel to its base, the upper and lower parts obtained are cone and a frustum respectively. 

8. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8 space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(A) 142296

(B) 142396

(C) 142496

(D) 142596

Solution

(A) 142296

According to the question,

Volume of cube =223=10648cm³ 

Volume of cube that remains unfilled =1/8×10648

=1331cm³ 

Volume occupied by spherical marbles =10648−1331

= 9317cm³ 

Radius of the spherical marble = 0.5/2

= 0.25cm

= 1/4cm

Volume of 1 spherical marble = (4/3) × (22/7) × (1/4)³ 
= 11/168 cm³ 

Numbers of spherical marbles, n = 9317 × (11/68)
= 142296

9. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

Solution

Volume of spherical shell = Volume of cone recast by melting

For Spherical Shell,

Internal diameter,

d₁ = 4 cm

Internal radius,

r₁ = 2 cm

External diameter,

d₂ = 8 cm

External radius,

r₂ = 4 cm

Now,
As volume of spherical shell = [4π/3(r₂³ - r₁³)]
Where r₁ and r₂ are internal and external radii respectively. 
Volume of given shell = [4π/3(4³ - 2³)]
= [(4π/3) × 56]
= (224/3)π
We know that, 

Volume of cone = (224/3)π cm³ 
For cone, 

Base diameter = 8 cm 

Base radius, r = 4 cm 

Let Height of cone = 'h'..

We know, 

Volume of cone = (1/3)π4² h
224π/3 = 16πh/3 
⇒ 16h = 224 
⇒ h = 14 cm 
So, Height of cone is 14 cm. 

10. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is molded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

Solution

AS we know, 

Volume of cuboid = lbh

Where,

l = length,

b = breadth and

h = height

For given cuboid,

Length,

l = 49 cm

Breadth,

b = 33 cm

Height,

h = 24 cm
Volume of cube  = 49 × 33 × 24 cm³ 
Now, Now

Let the radius of cube be r. 

As volume of sphere  = (4/3)πr³ Where r = radius of sphere 

Also, 

Volume of cuboid = volume of sphere molded 

49(33)(24) = (4/3)πr³ 
⇒ πr³ = 29106
⇒ r³ = 29106 × 22/7
⇒ r³ = 9261 = 21 cm 

Hence, radius of sphere is 21 cm 

11. A mason constructs a wall of dimensions 270cm× 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that 1/8 space is covered by the mortar. Then the number of bricks used to construct the wall is

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Solution

(B) 11200

Volume of wall = 270cm× 300cm × 350cm

=28350000 cm³ 
1/8 space is covered by the mortar, 

Remaining space of wall = Volume of wall - volume of mortar 
= 28350000 - 28350000 × 1/8 
= 24806250 
Now, 

Volume of 1 brick = 22.5 cm  × 11.25 cm × 8.75 cm 

= 2214.844 cm³ 
So, 

Required no. of bricks, 

= 24806250/2214.844
= 11200

12. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Solution

(c) Given,

Diameter of the cylinder = 2 cm

Radius = 1 cm and

Height of the cylinder = 16 cm  [diameter = 2 × radius]
Volume of the cylinder = π × (1)² × 16 
= 16π cm³ 
According to question, 

Volume of 12 solid spheres = Volume of cylinder 
⇒ 12 × (4/3)πr³ = 16π
⇒ r = 1 
Diameter of each sphere, d = 2r
= 2× 1 = 2 cm 
Hence, the required diameter of each sphere is  2 cm.

13. The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(A) 4950 cm² 
(B) 4951 cm² 

(C) 4952 cm² 

(D) 4953 cm² 

Solution

(a) Given, 

The radius of the top of the bucket, 

R = 28 cm 

The radius of the bottom of the bucket, 

r = 7 cm 

Slant height of the bucket, 

l = 45 cm 

Since, bucket is in the form of frustum of a cone. 

Curved surface area of the bucket = = πl(R + r)

= π × 45 (28 + 7)
= 4750 cm² 

14. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm³ 
(B) 0.35 cm³ 
(C) 0.34 cm³
(D) 0.33 cm³

Solution

(A) 0.36 cm³ 
Given, diameter of cylinder = Diameter of hemisphere = 0.5 cm 

Radius = 0.25 cm 

Total length of capsule = 2 cm 

So, 

Length of cylindrical part of capsule  = 2 - (0.25 + 0.25) = 1.5 cm 
So, 
Capacity of capsule = Volume of cylindrical part + 2 × volume of hemisphere 
= πr²h + (2/3)πr³ 

= 0.36 cm³ 

15. If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

(A) 4πr² 
(B) 6πr² 

(C) 3πr² 

(D) 8πr² 

Solution

(A) 4πr² 

Because curved surface area of a hemisphere is 2πr² and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere. 
Hence, the curved surface area of new solid = 2πr² + 2πr² = 4πr² 

16. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) 2h cm

Solution

(B) 2r cm

Because the sphere encloses in the cylinder, therefore the diameter of sphere is equal to

diameter of cylinder which is 2r cm.

17. During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

Solution

(C) remain unaltered

During conversion of a solid from one shape to another, the volume of the new shape will

remain unaltered.

18. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Solution

(A) 32.7 litres

Given,

Diameter of one end of the bucket = 44cm

Radius(R) = 22cm

Diameter of other end of the bucket = 24cm

Radius(r) = 12cm

h = 35cm

Capacity of bucket = Volume of frustum of cone
= (1/3)πh(R² + r² + rR) 
Putting value of R and r, we get, 

Capacity of bucket = 32.71 

19. In a right circular cone, the cross-section made by a plane parallel to the base is a

(A) circle

(B) frustum of a cone

(C) sphere

(D) hemisphere

Solution

(B) We know that, if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone.

20. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D) 16 : 9

Solution

(D) 16 : 9

Let the radii of the two spheres are r₁ and r₂ respectively. 

Therefore, 

Ratio of volume of spheres :