Maths MCQ Class IX Ch-13 | Surface Area & Volume

     Mathematics MCQ | Class 09 | Chapter 13

SURFACE AREA & VOLUME

Multiple Choice Questions (MCQ)

• MCQ Based on the different types of Solid Figures.
• MCQ  Based on the concept of Cube & Cuboid.
• MCQ Based on the Right circular Cylinder and Cone.
• MCQ Based on the Sphere & Hemi-Sphere.

Features

• In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations. 
• Solution Hints are also given to some difficult problems. 
• Each MCQ contains four options from which one option is correct. 

Action Plan

• First of all students should Learn and write all basic points and Formulas related to the Chapter 13 Surface Area and Volume.
• Start solving  the NCERT Problems with examples.
• Solve the important assignments on the Chapter 13 Class IX.
• Then start solving the following MCQ.

Complete Mathematics Formulas on Surface Area and Volume

MCQ BASED ON CUBE AND CUBOID

Cube and cuboid have following properties

Both have 6 faces, 8 vertices and 12 edges.

Faces of cube are squares and faces of cube are rectangles.

Q1) 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

Q2) If the perimeter of one of the faces of a cube is 40 cm, then its volume is:
a) 6000 cm³
b) 1600 cm³
c) 1000 cm³
d) 600 cm³

Answer: c

Q3) 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 = 6a²

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

Q4) 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

Q5) 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.

Q6) The length of the longest pole that can be put in a room of dimension (10 m × 10 m × 5 m) is
a) 15 m
b) 16 m
c) 10 m
d) 12 m

Answer: a

Q7) The total surface area of a cube is 96 cm². The volume of the cube is:

a) 8 cm³ 

b) 512 cm³

c) 64 cm³

d) 27 cm³

Answer: c

Explanation:

We know that the TSA of the cone = 6a².

6a² = 96 cm²

a² = 96/6 = 16

a = 4 cm

The volume of cone = a³ cubic units

V = 4³ = 64cm³.

Q8) The lateral surface area of a cube is 256 m³. The volume of the cube is

a) 512 m³

b) 64 m³

c) 216 m³

d) 256 m³

Answer: a

Q9) 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 = √[10² + 10² + 5²]

= √(100 + 100 + 25) = √225 = 15 m.

Q10) 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 = 400000cm³

Volume of pits = 1600cm × 1200cm × 400cm = 768000000cm³

Number of planks = Volume of planks/Volume of pits

= 768000000/400000

Hence, the number of pits = 1920

Q11) The total surface area of a cube is 96 cm². The volume of the cube is

a) 8 cm³

b) 512 cm³

c) 64 cm³

d) 27 cm³

Answer: c

Q12) The lateral surface area of a cube is 256 m². The volume of the cube is

a) 512 m³

b) 64 m³

c) 216 m³

d) 256 m³

Answer: a

Explanation:

The lateral surface area of cube = 4a²

4a² = 256

a² = 256/4 =64

a = 8 m

Hence, the volume of cube = a³ cube units 

V = 8³  = 512 m³.

MCQ BASED ON RIGHT CIRCULAR CYLLINDER

Cylinder is a solid figure which have three faces, two faces at the top and bottom are plan surface and one is curved surface or Lateral surface.

Q13) 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)

= (2 x 22 x 4 x 14)/7

= (2 x 22x 4 x 2)

= 352 sq.cm

Q14) 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) 3 : 1
c) 4 : 1
d) 1 : 8

Answer: c

Q15) 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”.

Q16) The curved surface area of a right circular cylinder of height 14 cm is 88 cm². The diameter of the base is:

a) 2 cm

b) 3cm

c) 4cm

d) 6cm

Answer: a

Q17) The curved surface area of a right circular cylinder of height 14 cm is 88 cm². 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

Q18) Volume of hollow cylinder is

a) π(R² – r²)h

b) πR²h

c) πr²h

d) πr²(h₁ – h₁)

Answer: a

Q19) 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

Q20) 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

Q 21) 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, r₁= 2r, r₂ = 3r

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

Hence, h₁ = 5h, h₂ = 3h

The ratio of the volume of two cylinders = V₁/V₂

= πr₁²h₁/πr₂²h₂

= [(2r)²(5h)]/[(3r)²(3h)]

Ratio of their volumes = (20r²h)/(27r²h) = 20/27 = 20 : 27.

Q 22) If the radius of a cylinder is 4cm and height is 10cm, then total surface area of cylinder is:

a) 440 sq.cm

b) 352 sq.cm.

c) 400 sq.cm

d) 412 sq.cm

Answer: b

MCQ BASED ON RIGHT CIRCULAR CONE

Cone is a solid figure which has two faces one is plane surface at the bottom and another is a curved surface.

Q 23) 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

Q 24) A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a shape. The radius of the sphere is

a) 4.2 cm

b) 2.1 cm

c) 2.4 cm

d) 1.6 cm

Answer b

Q 25) 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

Q 26) 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)πr²h = (4/3)πr³

(⅓)π(2.1)²(8.4) = (4/3)πr³

37.044 = 4r³

r³= 37.044/4

r³= 9.261

r = 2.1

Therefore, the radius of the sphere is 2.1 cm.

Q27) 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

Q28) 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)]

MCQ BASED ON SPHERE AND HEMI-SPHERE

Sphere is solid figure which has only one curved surface.

Hemi sphere is a solid figure which has two faces, one is plane surface at the top and another is curved surface.

Q29) The volume of a hemisphere whose radius is r is:

a) 4/3 πr³

b) 4πr³

c) 2πr³

d) 2/3 π r³

Answer: d

Q30) 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 = 4πr²

= 4 x 22/7 x (14)² = 2464 sq.cm.

Q 31)The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios 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

Q32) The radius of a sphere is 2r, then its volume will be

a) (4/3) πr³

b) 4πr³

c) (8/3) πr³

d) (32/3) πr³

Answer: d

Explanation:

Given : r = 2r

The volume of sphere = (4/3)πr³ = (4/3)π(2r)³

V = (4/3)π(8r³) = (32/3)πr³.

Q 33) 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 = 3πr² 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.

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