# Mathematics Questions and Answers – Surface Area and Volume of Combination of Solids – 2

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This set of Mathematics Assessment Questions for Class 10 focuses on “Surface Area and Volume of Combination of Solids – 2”.

1. Find the surface area of the given solid which is in the form of a cone mounted on a hemisphere. The radius and height of the cone are 5cm and 12cm.
a) 214.4 cm2
b) 279.53 cm2
c) 70 cm2
d) 72.5 cm2

Explanation: Slant height = $$\sqrt {h^2+r^2}$$
= $$\sqrt {12^2+5^2}$$
= √169
= 13cm
The surface area of the toy = C.S.A of the cone + C.S.A of the sphere
= πrl + 2πr2
= (3.14 × 3 × 13) + (2 × 3.14 × 52)
= 47.1 + 56.52
= 279.53cm2

2. A wooden box is in the shape of a cuboid with three conical depressions. 7 cm × 3 cm × 4 cm are the dimensions of the cuboid and the radius and depth of the conical depressions are 0.5 cm and 1.2 cm. Find the volume of the entire wooden box?
a) 109.4 cm3
b) 80.05 cm3
c) 150 cm3
d) 89.4 cm3

Explanation: Volume of the wooden box = volume of the cuboid – the volume of 3 conical depressions
= lbh – 3($$\frac {1}{3}$$πr2h)
= (7 × 3 × 4) – 3($$\frac {1}{3}$$ × 3.14 × 0.52 × 1.2)
= 80.05 cm3

3. Find the volume of the largest right circular cone that can be cut out of cube having 5 cm as its length of the side.
a) 32.72 cm3
b) 15 cm3
c) 25.4 cm3
d) 37.2 cm3

Explanation: Length of the side of the cube = height of the cone = 5 cm
The radius of the base of the cone = $$\frac {5}{2}$$ cm = 2.5 cm
The volume of the cone = $$\frac {1}{3}$$πr2h
= $$\frac {1}{3}$$ × 3.14 × 2.52 × 5
= 32.72 cm3

4. A toy is in the form of a cone mounted on a hemisphere and a cylinder. The radius and height of the cone are 3 m and 4 m. Find the volume of the given solid?
a) 193.21 m3
b) 207.30 m3
c) 184.21 m3
d) 282.21 m3

Explanation: Volume of the toy = volume of the cone + volume of the hemisphere + volume of the cylinder
= $$\frac {1}{3}$$πr2h + $$\frac {2}{3}$$πr3 + πr2h
= ($$\frac {1}{3}$$ × 3.14 × 32 × 4) + ($$\frac {2}{3}$$ × 3.14 × 33) + (3.14 × 32 × 4)
= 207.30 m3

5. What is the length of the resulting solid if two identical cubes of side 7 cm are joined end to end?
a) 26 cm
b) 16 cm
c) 21 cm
d) 14 cm

Explanation: Length of resulting cuboid = 2 × side of the cube
= 2 × 7 cm
= 14 cm

6. The length, breadth and height of the cuboid is 8 cm, 4 cm and 4 cm. Find the volume of the cuboid?
a) 152.76 cm3
b) 154 cm3
c) 128 cm3
d) 141.76 cm3

Explanation: The volume of the cuboid = lbh
= 8 × 4 × 4
= 128 cm3

7. What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder?
a) πr2h + 2(2πr3)
b) 2πrh – 2(πr2)
c) 2πrh + 2($$\frac {2}{3}$$ πr2)
d) πr2h – 2($$\frac {2}{3}$$πr3)

Explanation: Volume of the article = volume of the cylinder – 2(volume of the hemisphere)
= πr2h – 2($$\frac {2}{3}$$πr3)

8. What is the formula to find the height of an iron pillar consisting of a cylinder and a cone mounted on it?
a) The radius of the cylinder + 2(height of the cone)
b) Height of the cylinder + 2(height of the cone)
d) Height of the cylinder + height of the cone

Explanation: To find the height of an iron pillar consisting of a cylinder and a cone mounted on it, we require heights of both cone and cylinder.
Height of the pillar = height of the cylinder + height of the cone.

9. What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder where the radius, height of the cylinder is 5 cm, 8 cm respectively and the radius of the hemisphere is 5 cm?
a) 104.40 cm3
b) 205.6 cm3
c) 168.23 cm3
d) 604 cm3

Explanation: Volume of the article = volume of the cylinder – 2(volume of the hemisphere)
= πr2h – 2($$\frac {2}{3}$$πr3)
= (3.14 × 52 × 8) – 2($$\frac {2}{3}$$ × 3.14 × 53)
= 104.40 cm3

10. What is the formula required to use for T.S.A of an article which is made by digging out a hemisphere from each end of a solid cylinder?
a) C.S.A of the cylinder – 2(C.S.A of the hemisphere)
b) C.S.A of the cylinder + C.S.A of the hemisphere
c) C.S.A of the cylinder + 2(C.S.A of the hemisphere)
d) C.S.A of the cylinder – C.S.A of the hemisphere 