# Class 10 Maths Chapter 13 Surface Areas and Volumes MCQs

**Class 10 Maths Chapter 13 MCQs (Surface areas and volumes)** are provided here online, along with their solutions. Students who are preparing for the board exams can refer to these multiple-choice questions to score good marks in the final examination. These objective questions are framed as per the CBSE syllabus (2021-2022) and NCERT curriculum. To get all class 10 Maths chapter-wise MCQs, Click here.

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## Class 10 Maths MCQs for Surface Areas and Volumes

**1. The shape of an ice-cream cone is a combination of:**

(a) Sphere + cylinder

(b) Sphere + cone

(c) Hemisphere + cylinder

(d) Hemisphere + cone

Answer: **(d) Hemisphere + cone**

The shape of an ice-cream cone is a combination of a hemisphere and a cone.

**2. If a cone is cut parallel to the base of it by a plane in two parts, then the shape of the top of the cone will be a:**

(a) Sphere

(b) Cube

(c) Cone itself

(d) Cylinder

Answer: **(c) Cone itself**

Explanation: If we cut a cone into two parts parallel to the base, then the shape of the upper part remains the same.

**3. If we cut a cone in two parts by a plane parallel to the base, then the bottom part left over is the:**

(a) Cone

(b) Frustum of cone

(c) Sphere

(d) Cylinder

Answer:** (b) Frustum of cone**

Explanation: See the figure below

**4. If r is the radius of the sphere, then the surface area of the sphere is given by;**

(a) 4 π r^{2}

(b) 2 π r^{2}

(c) π r^{2}

(d) 4/3 π r^{2}

Answer: (a) 4 π r^{2}

If r is the radius of the sphere, then the surface area of the sphere is given by 4 π r^{2}.

**5. If we change the shape of an object from a sphere to a cylinder, then the volume of cylinder will**

(a) Increase

(b) Decrease

(c) Remains unchanged

(d) Doubles

Answer: **(c) Remains unchanged**

Explanation: If we change the shape of a three-dimensional object, the volume of the new shape will be same.

**6. Fifteen solid spheres are made by melting a solid metallic cone of base diameter 2cm and height 15cm. The radius of each sphere is:**

(a) ½

(b) ¼

(c) 1/^{3}√2

(d) 1/^{3}√4

Answer: **(d) 1/ ^{3}√4**

Explanation: Volume of 15 spheres = Volume of a cone

15 x (4/3) π r^{3 }= ⅓ πr^{2}h

5×4 π r^{3 }= ⅓ π 1^{2}(15)

20r^{3 }= 5

r^{3 }= 5/20 = ¼

r = 1/^{3}√4

**7. The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8 cm. The curved surface of the bucket is:**

(a) 4000 sq.cm

(b) 3500 sq.cm

(c) 3630 sq.cm

(d) 3750 sq.cm

Answer: **(c) 3630 sq.cm**

Explanation: Curved surface of bucket = π(R_{1 }+ R_{2}) x slant height (l)

Curved Surface = (22/7) x (25 + 8) x 35

CSA = 22 x 33 x 5 = 3630 sq.cm.

**8. If a cylinder is covered by two hemispheres shaped lid of equal shape, then the total curved surface area of the new object will be**

(a) 4πrh + 2πr^{2}

(b) 4πrh – 2πr^{2}

(c) 2πrh + 4πr^{2}

(d) 2πrh + 4πr

Answer: **(c) 2πrh + 4πr ^{2}**

Explanation: Curved surface area of cylinder = 2πrh

The curved surface area of hemisphere = 2πr^{2}

Here, we have two hemispheres.

So, total curved surface area = 2πrh + 2(2πr^{2}) = 2πrh + 4πr^{2}

**9. A tank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30 cm. The total surface area of the tank is:**

(a) 30 m

(b) 3.3 m

(c) 30.3 m

(d) 3300 m

Answer: **(b) 3.3 m**

Explanation: Total surface area of tank = CSA of cylinder + CSA of hemisphere

= 2πrh + 2πr^{2}= 2π r(h + r)

= 2 x 22/7 x 30(145 + 30) cm^{2}

=33000 cm^{2}

= 3.3 m^{2}

**10. If we join two hemispheres of same radius along their bases, then we get a;**

(a) Cone

(b) Cylinder

(c) Sphere

(d) Cuboid

Answer: **(c) Sphere**

If we join two hemispheres of same radius along their bases, then we get a Sphere.

**11. 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

Answer:** (a) a cone and a cylinder**

A cylindrical pencil sharpened at one edge is the combination of a cone and a cylinder.

**12. 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

Answer: **(d) frustum of a cone and a hemisphere**

A shuttle cock used for playing badminton has the shape of the combination of frustum of a cone and a hemisphere.

**13. A hollow cube of internal edge 22 cm 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

Answer: **(a) 142296 **

Explanation:

Volume of cube = 223 = 10648 cm^{3}

Volume of cube that remains unfilled = (1/8) × 10648 = 1331 cm^{3}

volume occupied by spherical marbles = 10648 − 1331 = 9317 cm^{3}

Radius of the spherical marble = 0.5/2 = 0.25 cm = 1/4 cm

Volume of 1 spherical marble = (4/3) × (22/7) × (1/4)3 = 11/168 cm^{3}

Numbers of spherical marbles = n = 9317 × (11/168) = 142296

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

(a) 21 cm

(b) 23 cm

(c) 25 cm

(d) 19 cm

Answer: **(a) 21 cm**

Explanation:

For the given cuboid,

Length, l = 49 cm

Breadth, b = 33 cm

Height, h = 24 cm

Volume of cube = 49 × 33 × 24 cm3

Let r be the radius of the sphere.

Volume of sphere = 4/3 πr^{3}

Volume of cuboid = volume of sphere molded

49 × 33 × 24 = 4/3 πr^{3}

⇒ πr^{3} = 29106

⇒ r^{3} = 29106 × (22/7)

⇒ r^{3} = 9261

⇒ r^{3} = (21)^{3}

⇒ r = 21 cm

Hence, the radius of sphere is 21 cm

**15. 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

Answer: **(b) 2r cm**

Explanation:

Radius of cylinder = r

Height = h > 2r

Since the sphere is enclosed by a cylinder, the radius of the cylinder will be the radius of the sphere.

Therefore, diameter of sphere = 2r cm

**16. 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

Answer: **(a) 32.7 litres **

Explanation:

Given,

The height of the bucket = h = 35 cm

Diameter of one circular end of bucket = 44 cm

Then the radius R = 22 cm

Diameter of another end = 24 cm

Then the radius r = 12 cm

We know that Volume of the bucket = (1/3)πh[R^{2} + r^{2} + Rr]

= (1/3) × (22/7) × 35 × [(22)^{2} + (12)^{2} + 22 × 12]

= (35/3) × (22/7) × (484 + 144 + 264)

= (5 × 22 × 892)/3

= 32706.6 cm^{3}

= 32.7 litres.

**17. Two identical solid cubes of side a are joined end to end. Then the total surface area of the resulting cuboid is**

(a) 12a^{2}

(b) 10a^{2}

(c) 8a^{2}

(d) 11a^{2}

Answer:** (b) 10a ^{2}**

Explanation:

The total surface area of a cube having side a = 6a^{2}

If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a^{2}.

**18. A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is **

(a) 4πrh + 4πr^{2}

(b) 2πrh + 4πr^{2}

(c) 2πrh + 2πr^{2}

(d) 4πrh + 2πr^{2}

Answer: **(d) 4πrh + 2πr ^{2}**

Explanation:

We know that,

The total surface area of cylinder = 2πrh + 2πr^{2}

When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder will be 2h and radius will be r.

Thus, the total surface area of the shape so formed = 2πr(2h) + 2πr^{2} = 4πrh + 2πr^{2}

**19. The number of shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm is approximately equal to**

(a) 84

(b) 90

(c) 92

(d) 80

Answer: **(a) 84**

Explanation:

Volume of cuboidal lead solid = 9 cm × 11 cm × 12 cm = 1188 cm^{3}

Radius of lead shot = 3/2 cm = 1.5 cm

Volume of each shot = (4/3)πr^{3}

= (4/3) × (22/7) × 1.5 × 1.5 × 1.5

= 14.143 cm^{3}

Number of lead shots can be made = 1188/14.143 = 84 (approx.)

**20. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is **

(a) 6πr^{2}

(b) 5πr^{2}

(c) 4πr^{2}

(d) 3πr^{2}

Answer: **(c) 4πr ^{2}**

Explanation:

When two hemispheres are joined together along their bases, a sphere of the same base radius is formed.

Curved Surface Area of a sphere = 4πr^{2}

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