# Class 9 Maths Chapter 13 Surface Areas and Volumes MCQs

Class 9 Maths Chapter 13 Surface areas and volumes MCQs are provided here with solutions. All the objective questions are based on the CBSE syllabus and the latest NCERT curriculum. Students are advised to practise these chapter-wise questions to prepare for final exams to score well. Also, check Important Questions for Class 9 Maths.

## MCQs on Class 9 Maths Chapter 13 Surface Areas and Volumes

Students of 9th Standard can choose the correct option and solve the MCQs on Surface areas and volumes.

**1) The formula to find the surface area of a cuboid of length (l), breadth (b) and height (h) is:**

a. lb+bh+hl

b. 2(lb+bh+hl)

c. 2(lbh)

d. lbh/2

Answer:** b**

**2) The surface area of a cube whose edge equals to 3cm is:**

a. 62 sq.cm

b. 30 sq.cm

c. 54 sq.cm

d. 90 sq.cm

Answer:** c**

Explanation: Given, a = 3 cm

Surface area of cube = 6a^{2}

SA = 6 x 3 x 3 = 54 sq.cm

**3) The surface area of cuboid-shaped box having length=80 cm, breadth=40cm and height=20cm is:**

a. 11200 sq.cm

b. 13000 sq.cm

c. 13400 sq.cm

d. 12000 sq.cm

Answer:** a**

Explanation: surface area of the box = 2(lb + bh + hl)

S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]

= 2[3200 + 800 + 1600]

= 2 × 5600 = 11200 sq.cm.

**4) The volume of a hemisphere whose radius is r is:**

a. 4/3 πr^{3}

b. 4πr^{3}

c. 2πr^{3}

d. ⅔ π r^{3}

Answer:** d**

**5) If the radius of a cylinder is 4cm and height is 10cm, then the total surface area of a cylinder is:**

a. 440 sq.cm

b. 352 sq.cm.

c. 400 sq.cm

d. 412 sq.cm

Answer:** b**

Explanation: Total Surface Area of a Cylinder = 2πr(r + h)

TSA = 2 x 22/7 x 4(4 + 10)

= (2x22x4x14)/7

= (2x22x4x2)

= 352 sq.cm

**6) The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. The diameter of the base is:**

a. 2 cm

b. 3cm

c. 4cm

d. 6cm

Answer:** a**

Explanation: Curved surface area of cylinder = 88 sq.cm

Height = 14 cm

2πrh = 88

r = 88/2πh

r=1 cm

Diameter = 2r = 2cm

**7) The Curved surface area of a right circular cylinder is 4.4 sq.cm. The radius of the base is 0.7 cm. The height of the cylinder will be:**

a. 2 cm

b. 3 cm

c. 1 cm

d. 1.5 cm

Answer:** c**

Explanation: Curved surface area of cylinder = 2πrh

2πrh = 4.4

h = 4.4/(2π x 0.7)

h = 1 cm

**8) The diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. The curved surface area is:**

a. 150 sq.cm

b. 165 sq.cm

c. 177 sq.cm

d. 180 sq.cm

Answer:** b**

Explanation: Diameter = 10.5, Radius = 10.5/2

Slant height, l = 10cm

Curved surface area of cone = πrl = π(5.25)(10)

CSA = 165 sq.cm

**9) If slant height of the cone is 21cm and the diameter of the base is 24 cm. The total surface area of a cone is:**

a. 1200.77 sq.cm

b. 1177 sq.cm

c. 1222.77 sq.cm

d. 1244.57 sq.cm

Answer:** d**

Explanation: Total surface area = πr(l + r)

r = 24/2 = 12 cm

l = 21 cm

TSA = π(12)(21 + 12) = 1244.57 sq.cm

**10) The surface area of a sphere of radius 14 cm is:**

a. 1386 sq.cm

b. 1400 sq.cm

c. 2464 sq.cm

d. 2000 sq.cm

Answer:** c**

Explanation: Radius of sphere, r = 14 cm

Surface area = 4πr^{2}

= 4 x 22/7 x (14)^{2} = 2464 sq.cm.

**11) The radius of a sphere is 2r, then its volume will be**

(a) (4/3) πr^{3}

(b) 4πr^{3}

(c) (8/3) πr^{3}

(d) (32/3) πr^{3}

Answer: **d**

Explanation:

Given : r=2r

The volume of sphere = (4/3)πr^{3 }= (4/3)π(2r)^{3}

V = (4/3)π(8r^{3}) = (32/3)πr^{3}.

**12) The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is**

(a) 1:4

(b) 1:3

(c) 2:3

(d) 2:1

Answer: **a**

Explanation:

We know that the total surface area of the hemisphere = 3πr^{2} square units.

If r= 6cm, then TSA = 3π(6)^{2} = 108π

If r = 12 cm, then TSA = 3π(12)^{2}= 432π

Then the ratio = (108π)/(432π)

Ratio = ¼, which is equal to 1:4.

**13) The total surface area of a cube is 96 cm**^{2}**. The volume of the cube is:**

(a) 8 cm^{3}

(b) 512 cm^{3}

(c) 64 cm^{3}

(d) 27 cm^{3}

Answer: **c**

Explanation:

We know that the TSA of the cone = 6a^{2}.

6a^{2} = 96 cm^{2}

a^{2} = 96/6 = 16

a =4 cm

The volume of cone = a^{3} cubic units

V = 4^{3} = 64cm^{3}.

**14) The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5m) is**

(a) 15m

(b) 16m

(c) 10m

(d) 12m

Answer: **a**

Explanation:

Given: l=10m, b= 10m, h= 5m

The length of the longest pole = √[10^{2}+10^{2}+5^{2}]

= √(100+100+25) = √225 = 15 m.

**15) A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is**

(a) 4.2 cm

(b) 2.1 cm

(c) 2. 4 cm

(d) 1.6 cm

Answer: **b**

Explanation:

Given that the height of cone = 8.4 cm

Radius of cone = 2.1 cm

Also, given that the volume of cone = volume of a sphere

(⅓)πr^{2}h = (4/3)πr^{3}

(⅓)π(2.1)^{2}(8.4) = (4/3)πr^{3}

37.044= 4r^{3}

r^{3}= 37.044/4

r^{3}= 9.261

r = 2.1

Therefore, the radius of the sphere is 2.1 cm.

**16) The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit that is 16 m long, 12m wide and 4 m deep is**

(a) 1900

(b) 1920

(c) 1800

(d) 1840

Answer: **b**

Explanation:

Volume of Plank = 400 cm×50cm×20cm=400000cm^{3}

Volume of pits = 1600cm×1200cm×400cm = 768000000cm^{3}

Number of planks = Volume of planks/Volume of pits

= 768000000/400000

Hence, the number of pits = 1920

**17) In a cylinder, the radius is doubled and height is halved, the curved surface area will be**

(a) Halved

(b) Doubled

(c) Same

(d) Four times

Answer: c

Explanation:

We know that the curved surface area of a cylinder is 2πrh

Given that, r = 2R, h= H/2

Hence, the CSA of new cylinder = 2π(2R)(H/2) = 2πRH

Therefore, the answer is “Same”.

**18) The lateral surface area of a cube is 256 m**^{2}**. The volume of the cube is**

(a) 512 m^{3}

(b) 64 m^{3}

(c) 216 m^{3}

(d) 256 m^{3}

Answer: **a**

Explanation:

The lateral surface area of cube = 4a^{2}

4a^{2}= 256

a^{2} = 256/4 =64

a = 8 m

Hence, the volume of cube = a^{3} cube units

V = 8^{3 } = 512 m^{3}.

**19) The total surface area of a cone whose radius is r/2 and slant height 2l is**

(a) 2πr(l+r)

(b) πr(l+(r/4))

(c) πr(l+r)

(d) 2πrl

Answer: **b**

Explanation:

The total surface area of cone = πr(l+r) square units.

If r = r/2 and l= 2l, then the TSA of cone becomes,

TSA of cone = π(r/2)[(2l+(2/r)]

=π[(rl)+(r2/4)]

TSA of new cone =πr[l+(r/4)]

**20) The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3. The ratio of their volumes is:**

(a) 10: 17

(b) 20: 27

(c) 17: 27

(d) 20: 37

Answer: **b**

Explanation:

Given that, the radii of two cylinders are in the ratio of 2:3

Hence, r_{1}= 2r, r_{2} = 3r

Also, given that, the height of two cylinders are in the ratio 5:3.

Hence, h_{1} = 5h, h_{2} = 3h

The ratio of the volume of two cylinders = V_{1}/V_{2}

= πr_{1}^{2}h_{1}/πr_{2}^{2}h_{2}

= [(2r)^{2}(5h)]/[(3r)^{2}(3h)]

Ratio of their volumes =(20r^{2}h)/(27r^{2}h) = 20/27 = 20: 27.

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