Surface Area And Volume Chapter 12: NCERT Exemplar for Class 10

NCERT Exemplar Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes are provided here in PDF format so that students can access them and prepare well for the CBSE Board exams. These exemplar problems and solutions are designed by Maths experts with respect to the latest syllabus of CBSE (2024-2025).

Students at simply acad can download the PDF of these solved questions for free and use them to get a clear idea about the importance and formulas as well as practise questions and to develop better Math skills. They can also use this NCERT Exemplar to solve even the difficult questions and be ready for the board exams. Moreover, the sample papers and previous years’ question papers are of great help in getting familiar with the types of questions asked in the exam from Chapter 12 and the marking scheme. This chapter covers the following topics of Surface Areas and Volumes:simply acad not only provides study material but also create the content in such a way that can be easily understood and evolve in the mind of students in order to avoid any difficulty in,during and before the board examinations as well.

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Question 1: A cylindrical pencil sharpened at the one edge has 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

Answer 1: (A) The cone and the cylinder

Explanation: A pencil has the combination of

The nip of the sharpened pencil is in a conical shape, and the rest of the part is cylindrical; hence the pencil is the combination of a cylinder and a cone.

So, the correct answer is option (A).

 

Question 2: A surahi has the combination of

(A) a sphere as well as a cylinder

(B) a hemisphere as well as a cylinder

(C) the two hemispheres

(D) a cylinder as well as a cone.

Answer 2: (A) a sphere as well as a cylinder

Explanation: The top part of the surahi is in a cylindrical shape and the bottom part is in a spherical shape; hence surahi is a combination of a Sphere and a cylinder.

So the correct answer is option (A).

 

Question 3: A plumbline (Sahul) has the combination of (see Fig. )

(A) a cone as well as a cylinder

(B) a hemisphere as well as a cone

(C) frustum of the cone as well as a cylinder

(D) the sphere as well as cylinder

Answer 3: (B) a hemisphere as well as a cone

Explanation: The upper part of the plumbline is hemispherical, and the bottom part is conical. Hence, it is a combination of a hemisphere as well as a cone.

Hence, the correct answer is option (B).

 

Question 4:The shape for a glass (tumbler) (see Fig. ) is usually of the form of

(A) The cone

(B) frustum of the cone

(C) the cylinder

(D) the Sphere

Answer 4: (B) frustum of the cone

Explanation: The shape of the frustum of the cone is:

Hence, the shape for glass is a frustum [an inverted frustum].

Here, the correct answer is option (B).

 

Question 5: The shape for a Gilli, in the Gilli-danda game (see Fig.), has a combination of

(A) the two cylinders

(B) a cone as well as a cylinder

(C) two cones as well as a cylinder

(D) two cylinders as well as a cone

Answer 5: (C) two cones as well as a cylinder

Explanation: A Gilli has a combination of

The left and the right part for a Gilli are conical and the central part is cylindrical.

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

So, the correct answer is option (C).

 

Question 6:A shuttle cock used for playing badminton has the shape for the combination of

(A) a cylinder as well as a sphere

(B) a cylinder as well as a hemisphere

(C) a sphere as well as a cone

(D) frustum of the cone as well as a hemisphere

Answer 6: (D) frustum of the cone as well as a hemisphere

Explanation: The cork of the shuttle is hemispherical in shape as well as the upper part is in the shape of the frustum of a cone. Hence, it is a combination of frustum of a cone and a hemisphere.

Thus, the correct answer is Option (D).

 

Question 7: A cone intersects through a plane parallel to its base as well as then the cone that is formed at one side for the plane is removed. The new part which is left over on the other side of the plane is called.

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Answer 7:(A) a frustum of a cone

Explanation: if a cone is divided with two parts with a plane through any point on the axis parallel to the base, the upper and lower parts obtained are cones as well as a frustum, respectively.

So, the correct answer is option (A).

 

Question 8: A hollow cube with an internal edge of 22 cm is filled with spherical marbles of the diameter of 0.5 cm as well as it is assumed that 1/8 of space for the cube remains unfilled. The number of marbles which the cube can use is

(A) 142244

(B) 142344

(C) 142444

(D) 142544

Answer 8: (A) 142244

Explanation:

Diameter of the marble = 0.5 cm

Radius = .5/2 = 5/ 20 = 1/ 4 cm

volume of the marble =(4πr3)/3 = 4/3 *22/7 * ¼ * ¼ * ¼ = 11/168 cm3

Edge for the cube = 22 cm

Volume (V)= 22 * 22 * 22

Space occupied with marble = total volume 1/8 part of the volume = v – v/8 = 7v/8

Number of the marbles = space occupied/ volume of marble

=(7v * 168)/ 8*11

= (7*22*22*22*22*168)/(8*11)

= 142244

 

Question 9: A metallic spherical shell for internal and external diameters 4 cm and 8 cm, respectively, is melted as well as recast to form a cone of the base diameter of 8 cm. The height of the cone is

(A) 12 cm

(B) 14 cm

(C) 15 cm

(D) 18 cm

Answer 9: (B) 14cm

Explanation: volume of the spherical shell = volume of the cone recast by melting

In case of spherical Shell,

Internal diameter is d1 = 4 cm

Internal radius is r1 = 2 cm

[ as the radius = 1/2 diameter]

External diameter is d2 = 8 cm

External radius is r2 = 4 cm

Then,

When the volume of spherical shell= 4/3 π (r2 3 – r13)

here r1 and r2 are internal and external radii respectively.

volume of given shell = 4/3 π (43 – 23)

= 4/3 π (56)

= (224/3) π

We find

volume of the cone = 224π /3 cm3

In the case of cone,

For Base diameter = 8 cm

Base radius is r = 4 cm

Let height of the cone = ‘h’.

We get,

volume of the cone = (1/3) π r2h,

here r = Base radius and h = height of cone

volume of the given cone = (1/3) π 42h

224π /3 = 16πh /3

16h = 224

h = 14 cm

Thus, height of the cone is 14 cm.

 

Question 10: A solid piece for iron in the form of a cuboid having dimensions 49 cm × 33 cm × 24 cm is molded to form the solid Sphere. The radius of the Sphere is

(A) 21cm

(B) 23 cm

(C) 25 cm

(D) 19 cm

Answer 10: (A) 21cm

Explanation:

length of cuboid = 49 cm

Breadth for cuboid = 33 cm

height of cuboid = 24 cm

It is given by,

Volume = l * b * h

= 49*33*24 = 38808

We now know that volume of the sphere (4πr3)/3

For the volume of the sphere = 38808 (given)

r3 = (38808 *7*3)/(4*22) = 882*7*3)/2 = 9261

r = 3√(3*3*3*7*7*7) =3*7 = 21

Thus, the radius of the Sphere = 21 cm

 

Question 11: A mason constructs a wall for the dimensions 270 cm × 300 cm × 350 cm by the bricks, each having a size of 22.5 cm × 11.25 cm × 8.75 cm as well as it is assumed that 1/8 space is covered with the mortar. The number for bricks used to construct the wall is

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Answer 11: (B) 11200 bricks

Explanation:

length of the wall = 270 cm

Breadth for the wall = 300 cm

Height of the wall = 350 cm

We have,

Volume = l * b * h

=270 * 300 * 350 = 28350000cm3

length of the brick = 22.5 cm

Breadth for the brick = 11.25 cm

height of the brick = 8.75 cm

We get,

Volume = l * b * h

=22.5*11.25*8.75 = 2214.84375cm3

1/8 Space is covered with mortar (given)

The remaining space = volume of wall /8

2835000/8 = 3543750cm3

The surface constructed is = 28350000 – 3543750 = 24806250cm3

Number of the bricks used = surface contracted/volume of brick

=24806250/2214.84375 = 11200 bricks

 

Question 12: Twelve solid spheres for the same size are made with melting a solid metallic cylinder with a base diameter of 2 cm and a height of 16 cm. The Diameter of each Sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Answer 12: (C) 2 cm

Explanation:

Diameter of the metallic cylinder = 2 cm

Radius is = 2/2 = 1 cm

Height is = 16cm

We have,

Volume =πr2h

= π*1*1*16 = 16π

We are aware that the twelve solid sphere are made with melting of solid metallic cylinder

volume of the sphere = (4πr3)/3

So, the volume of the 12 spheres = 16π

= 12*(4πr3)/3 = 16π

= 16πr3 = 16π

r3 = 1

r = 1

Radius is = 1 cm

Diameter is 2*1 = 2 cm

 

Question 13: The radii for the top as well as the bottom of a bucket for the slant height of 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

Answer 13: (A) 4950 cm2

Explanation:

Slant height of the bucket = 45 cm

Top radius is = r1 = 28cm

Bottom radius is = r2 = 7cm +

Curved surface area of the bucket is = πl(r1+r2)

=22/7 * 45 * (28+7)

= 22/7 *45 *35

=4950cm2

 

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

(A) 0.36 cm3

(B) 0.35 cm3

(C) 0.34 cm3

(D) 0.33 cm3

Answer 14: (A) 0.36 cm3

Explanation:

Diameter of the hemisphere = 0.5 cm

Radius = 0.5/2 = 5/20 = ¼ = 0.25 cm

Volume of hemisphere = (2πr3)/3

=(2/3)*(22/7)*(1/4)*(1/4)*(1/4) =11/336

Volume of two hemispheres = (2*11)/336 = 11/168

Same as, the radius of the cylinder = 0.25

Height is = 2- 0.25 -0.25

= 2- 0.5

=1.5 cm

We get, volume = πr2h

= (22/7)*(1/4)*(1/4)*(15/10) = 33/112

The total volume of the capsule = volume of two hemispheres + volume of the cylinder

= (11/168)+(33/112)

=0.065 + 0.294

= 0.359cm3

=0.36cm3 (approximate)

 

Question 15: If two solid hemispheres for the same base radius r are joined together along with their bases, the curved surface area of this new solid is

(A) 4πr2

(B) 6πr2

(C) 3πr2

(D) 8πr2

Answer 15: (A) 4πr2

Explanation: The radius of the hemisphere = r

We know curved surface area = 2πr2

The curved surface area of two solid hemisphere

= 2 * 2πr2

= 4πr2

 

Question 16: A right circular cylinder having the radius r cm as well as height h cm (h > 2r) just encloses the Sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) Two h cm

Answer 16: (B) 2r cm

Explanation: Since the Sphere is just enclosed in a cylinder,

The Diameter of the Sphere will be equal to the diameter of the cylinder.

Diameter of the base of cylinder = 2(radius of base) = 2r

Thus, the Diameter of the Sphere = 2r cm

Therefore, the correct answer is option (B).

 

Question 17: During the conversion of the solid from one shape to another, the volume of the new shape will be

(A) increase

(B) decrease

(C) remains unaltered

(D) be doubled

Answer 17: (C) remains unaltered.

Explanation: When a solid is converted from one shape to another, the volume of the new shape remains the same.

Thus, the correct answer is option (C).

 

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

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Answer 18: (A) 32.7 litres

Explanation:

volume of the frustum of cone =(πh/3) *(r12+r22+r1r2)

Diameter of the first end = 44 cm

Radius is r1 = 44/2 = 22cm

Diameter of the second end = 24 cm

Radius is r2 = 24/2 = 12cm

For Height (h) = 35cm

Volume is = (πh/3) *(r12+r22+r1r2)

= 22/7 * 35/3 * (222+122+22*12)

=110/3(484+144+264)

=(110/3) * (892)

=32706.67cm3

We now know that 1 liter = 1000 cm3

Therefore, 32706.67cm3 = 32.7l

 

Question 19: In the right circular cone, the cross-section made with a plane parallel to the base is as follows:

(A) circle

(B) frustum of a cone

(C) sphere

(D) hemisphere

Answer 19: (B) frustum of a cone

Explanation: As per the question, when a right circular cone is cut with a plane parallel to the base, the figure formed is

Where BECD is not a circle, not a sphere, not a hemisphere but is instead a frustum of a cone.

Therefore, in a right circular cone, the cross-section made with a plane parallel to the base is a frustum of a cone.

 

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

(A) 3: 4

(B) 4 : 3

(C) 9: 16

(D) 16: 9

Answer 20: (D) 16 : 9

Explanation: Assume two-sphere having radius r1 and r2

As per the question,

volume of the first sphere / volume of the second sphere = 64/27

= (4/3 *πr13)/(4/3 *πr23) = 64/27

(r1/r2)3 = 64/27

r1/r2 = 3√(64/27) =4/3

Ratio for their surface area is = (4 *πr12)/(4 *πr22) = r12/r22 = (r1/r2)2 = (4 /3)2 = 16/9

So, the required ratio is 16: 9

 

Question 21. Two identical solid hemispheres for equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr2.

Answer 21: False

Explanation:

If two hemispheres are joined together along with their bases, a sphere for the same base radius is formed.

So, Curved Surface area of a sphere = 4πr2.

 

Question 22. A solid cylinder for the radius r and height h is placed over other cylinders of the same height and radius. The total surface area of the shape so formed is 4πrh + 4πr2.

Answer 22: False

Explanation:

As per the question,

If one cylinder is placed over another, the base for the first cylinder and the top for the other cylinder will not be covered in the total surface area.

We get,

Total surface area of the cylinder = 2πrh + 2πr2h, here r = base radius and h = height

Total surface area of the shape formed = 2(Total surface of single cylinder) – 2(Area of base of cylinder)

= 2(2πrh + 2πr2) – 2(πr2)

= 4πrh + 2πr2

Question 23. A solid ball is perfectly fitted inside the cubical box for the side a. The volume of the ball is 4/3πa3.

Answer 23:False

Explanation:

Assume the radius of the sphere = r

If a solid ball is exactly fitted inside the cubical box for the side a,

We find,

Diameter of the ball = Edge length of the cube

2r = a

For Radius, r = a/2

We get,

volume of the Sphere = 4/3πr3

volume of the ball

= 4/3π(a/2)3

= 4/3π(a3/8)

= 1/6πa3

 

Question 24. Two identical cubes, each having a volume of 64 cm3, are joined together end to end. What is the surface area of the resulting cuboid?

Answer 24:

Assume the side of one cube = a

Surfaces area of the resulting cuboid

= 2(Total surface area of a cube) – 2(area of a single surface)

We get,

Total surface area of the cube = 6a2 , here a = side of cube

Surfaces area of the resulting cuboid = 2(6a2) – 2(a2) = 10a2

And,

As per the question,

volume of the cube = 64 cm3

For the Volume of cube, we have = a3

64 = a3

a = 4 cm

Hence,

Surface area of the resulting cuboid

= 10a2

= 10(4)2

= 160 cm2

 

Question 25. From the solid cube for a side of 7 cm, a conical cavity for the height of 7 cm as well as a radius of 3 cm is hollowed out. Find out the volume of the remaining solid.

Answer 25:

From the above figure, we find,

volume of the remaining solid

= volume of the cube – volume of the cone

In case of Cube

Side is a = 7 cm

We get,

volume of the cube = a3, here a = side of cube

volume of the cube = (7)3 = 343 cm3

In case of cone

Radius is, r = 3 cm

Height is, h = 7 cm

volume of the cone

= 1/3 πr2h

= 1/3 π(3)27

= 3 × (22/7) × 7

= 66 cm3

volume of remaining solid

= volume of the cube – volume of the cone

= 343 – 66

= 277 cm3

 

Question 26. 2 cubes each for the volume of 64 cm3 are joined end to end. Find out the surface area of the resulting cuboid.

Answer 26:

The diagram is given as follows:

Given that,

The Volume (V) for each cube is = 64 cm3

It implies that a3 = 64 cm3

Hence, a = 4 cm

Then, the side for the cube = a = 4 cm

And the length and breadth for the resulting cuboid will be of 4 cm each whereas its height will be of 8 cm.

Hence, the surface area of the cuboid = 2(lb+bh+lh)

= 2(8×4+4×4+4×8) cm2

= 2(32+16+32) cm2

= (2×80) cm2 = 160 cm2

 

Question 27. A vessel is of the form for a hollow hemisphere mounted with a hollow cylinder. The Diameter of the hemisphere is 14 cm as well as the total height of the vessel is 13 cm. Find out the inner surface area of the vessel.

Answer 27:

The diagram given below is:

Then, the given parameters are as follows:

The Diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

And the height of the cylinder = h = (13-7) = 6 cm

Also, the radius of the hollow hemisphere = 7 cm

Then, the inner surface area of the vessel = CSA for the cylindrical part + CSA for the hemispherical part

(2πrh+2πr2) cm2 = 2πr(h+r) cm2

2×(22/7)×7(6+7) cm2 = 572 cm2

 

Question 28. A cubical block of the side 7 cm is surmounted with a hemisphere. What is the greatest diameter the hemisphere could have? Find out the surface area of the solid.

Answer 28:

It is given that each side for the cube is 7 cm. Hence, the radius will be 7/2 cm.

We get,

The total surface area of the solid (TSA)

= surface area of the cubical block + CSA for the hemisphere – area of the base of the hemisphere

So, TSA of the solid = 6×(side)2+2πr2-πr2

= 6×(side)2+πr2

= 6×(7)2+(22/7)×(7/2)×(7/2)

= (6×49)+(77/2)

= 294+38.5 = 332.5 cm2

Hence, the surface area of the solid is 332.5 cm2

 

Question 29. A hemispherical depression is cut off from one face for a cubical wooden block so that the diameter l for the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer 29:

The diagram is given below:

Then, the Diameter of the hemisphere = Edge of the cube = l

Hence, the radius of the hemisphere = l/2

Thus, The total surface area of the solid

= surface area of the cube + CSA for the hemisphere – area of the base of the hemisphere

TSA for the remaining solid

= 6 (edge)2+2πr2-πr2

= 6l2 + πr2

= 6l2+π(l/2)2

= 6l2+πl2/4

= l2/4(24+π) sq. units

 

Question 30. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each for its ends. The length of the entire capsule is 14 mm, and the Diameter of the capsule is 5 mm. Find out its surface area.

Answer 30:

Two hemispheres, as well as one cylinder, are shown in the figure given below.

Where the Diameter of the capsule = 5 mm

So, the Radius = 5/2 = 2.5 mm

Then, the length of the capsule = is 14 mm

Hence, the length of the cylinder = 14-(2.5+2.5) = 9 mm

Therefore, The surface area of the hemisphere = 2πr2 = 2×(22/7)×2.5×2.5

= 275/7 mm2

Then, the surface area of the cylinder = 2πrh

= 2×(22/7)×2.5×9

(22/7)×45 = 990/7 mm2

So, the required surface area of the medicine capsule will be

= 2×surface area of the hemisphere + surface area of the cylinder

= (2×275/7) × 990/7

= (550/7) + (990/7) = 1540/7 = 220 mm2