# Class 10 Maths MCQ – Conversion of Solid from One Shape to Another

This set of Class 10 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Conversion of Solid from One Shape to Another”.

1. A metallic sphere whose radius is 5 cm is melted and cast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
a) 21.4 cm
b) 43.63 cm
c) 70 cm
d) 72.5 cm2

Explanation: Volume of the sphere = volume of the cylinder
$$\frac {4}{3}$$πr3 = πr2h
$$\frac {4}{3}$$ × 3.14 × 53 = 3.14 × 62 × h
h = $$\frac {4 \times 3.14 \times 3.14 \times 125}{3.14 \times 36}$$
h = 43.63 cm

2. Three metallic spheres of radius 3 cm, 6 cm, 9 cm are melted into a single sphere. Find the radius of the resulting sphere.
a) 109.4 cm3
b) 4071.48 cm3
c) 1520 cm3
d) 1869.4 cm3

Explanation: Volume of the resulting sphere = sum of the volumes of all three spheres
= $$\frac {4}{3}$$π33 + $$\frac {4}{3}$$π63 + $$\frac {4}{3}$$π93
= 113.09 + 904.77 + 3053.62
= 4071.48 cm3

3. A well of depth 25 m with a radius 4 m is dug from the earth forming a platform of length 28 m and a breadth of 16 m. Find the height of the platform.
a) 2.8 m3
b) 5 m3
c) 5.4 m3
d) 7.2 m3

Explanation: Volume of the well = volume of the platform
πr2h = lbh
3.14 × 42 × 25 = 28 × 16 × h
h = 2.8 m3

4. How many coins of 1 cm in diameter and thickness of 1.2 cm need to be melted to form a cuboid with dimensions of 5 cm × 10 cm × 4 cm?
a) 193
b) 213
c) 184
d) 282

Explanation: Volume of the cuboid = number of coins(volume of a coin)
lbh = number of coins(πr2h)
5 × 10 × 4 = number of coins(3.14 × 0.52 × 1.2)
200 = number of coins(0.94)
number of coins = 213

5. A sphere of radius 14 cm is melted and cast into a number of tiny cones of radius 2.33 cm each and height 6 cm. Find the number of cones that will be formed?
a) 726
b) 816
c) 721
d) 821

Explanation: Number of cones(volume of a cone) = volume of sphere
number of cones($$\frac {1}{3}$$πr2h) = $$\frac {4}{3}$$πr3
number of cones($$\frac {1}{3}$$ × 3.14 × 2.332 × 6) = $$\frac {4}{3}$$ × 3.14 × 143
number of cones(14) = 11488.21
number of cones = 821
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6. How many cylinders having 2.1 cm of radius and 1.4 cm of height can be made out of a cuboid metal box having dimensions 33 cm, 21 cm, 10.5 cm?
a) 152
b) 154
c) 844
d) 841

Explanation: Number of cylinders(volume of a cylinder) = volume of a cuboid
Number of cylinders (πr2h) = lbh
Number of cylinders (3.14 × 1.42 × 1.4) = 33 × 21 × 10.5
Number of cylinders = 844

7. A sphere having a radius of 3 cm is melted and elongated into a wire having a circular cross-section of radius 0.1 cm. Find the length of the wire?
a) 2400 cm
b) 3100 cm
c) 1200 cm
d) 3600 cm

Explanation: Volume of the wire = volume of the sphere
πr2h = $$\frac {4}{3}$$πr3
3.14 × 0.12 × h = $$\frac {4}{3}$$ × 3.14 × 27
h = 3600 cm

8. A cylindrical hole of depth 20 m with a radius 5 m is dug from the earth forming a platform of length 14 m and a breadth of 12 m. Find the height of the platform.
a) 8.34 m3
b) 11.84 m3
c) 7.64 m3
d) 9.34 m3

Explanation: Volume of the cylindrical hole = volume of the platform
πr2h = lbh
3.14 × 52 × 20 = 14 × 12 × h
h = 9.34 m3

9. What is the formula to find the rise in the water level when ‘x’ spherical balls are dropped into a cylindrical beaker?
a) $$\frac {Volume \, of \, ‘x’ \, spherical \, balls}{Base \, area \, of \, cylinder}$$
b) $$\frac {Volume \, of \, ‘x’ \, spherical \, balls}{2(Base \, area \, of \, cylinder)}$$
c) The volume of the cylinder + volume of the beaker
d) The volume of the cylinder – 2(volume of the beaker)

Explanation: To find the rise in the water level we require the volume of ‘x’ spherical balls and the base area of the cylinder
Rise in the water level = $$\frac {Volume \, of \, ‘x’ \, spherical \, balls}{Base \, area \, of \, cylinder}$$

10. A metallic sphere whose radius is 4 cm is melted and cast into the shape of a right circular cone of radius 7 cm. Find the height of the cylinder?
a) 14.48 cm
b) 22.36 cm
c) 16.40 cm
d) 20.32 cm

Explanation: Volume of the sphere = volume of the cone
$$\frac {4}{3}$$ πr3 = $$\frac {1}{3}$$ πr2h
$$\frac {4}{3}$$ × 3.14 × 43 = 3.14 × 72 × h
h = $$\frac {4 \times 3.14 \times 3.14 \times 64}{3.14 \times 49}$$
h = 16.40 cm

11. Three metallic spheres of radius 2 cm, 4 cm, 8 cm are melted into a single sphere. Find the radius of the resulting sphere.
a) 1009.4 cm3
b) 2446.25 cm3
c) 1520 cm3
d) 2869.4 cm3

Explanation: Volume of the resulting sphere = sum of the volumes of all three spheres
= $$\frac {4}{3}$$π23 + $$\frac {4}{3}$$π43 + $$\frac {4}{3}$$π83
= $$\frac {4}{3}$$ × 3.14(23 + 43 + 83)
= 2446.25 cm3

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