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 cm^{2}

b) 279.53 cm^{2}

c) 70 cm^{2}

d) 72.5 cm^{2}

View Answer

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 + 2πr

^{2}

= (3.14 × 3 × 13) + (2 × 3.14 × 5

^{2})

= 47.1 + 56.52

= 279.53cm

^{2}

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 cm^{3}

b) 80.05 cm^{3}

c) 150 cm^{3}

d) 89.4 cm^{3}

View Answer

Explanation: Volume of the wooden box = volume of the cuboid – the volume of 3 conical depressions

= lbh – 3(\(\frac {1}{3}\)πr

^{2}h)

= (7 × 3 × 4) – 3(\(\frac {1}{3}\) × 3.14 × 0.5

^{2}× 1.2)

= 80.05 cm

^{3}

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 cm^{3}

b) 15 cm^{3}

c) 25.4 cm^{3}

d) 37.2 cm^{3}

View Answer

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 = \(\frac {1}{3}\)πr

^{2}h

= \(\frac {1}{3}\) × 3.14 × 2.5

^{2}× 5

= 32.72 cm

^{3}

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 m^{3}

b) 207.30 m^{3}

c) 184.21 m^{3}

d) 282.21 m^{3}

View Answer

Explanation: Volume of the toy = volume of the cone + volume of the hemisphere + volume of the cylinder

= \(\frac {1}{3}\)πr

^{2}h + \(\frac {2}{3}\)πr

^{3}+ πr

^{2}h

= (\(\frac {1}{3}\) × 3.14 × 3

^{2}× 4) + (\(\frac {2}{3}\) × 3.14 × 3

^{3}) + (3.14 × 3

^{2}× 4)

= 207.30 m

^{3}

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

View Answer

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 cm^{3}

b) 154 cm^{3}

c) 128 cm^{3}

d) 141.76 cm^{3}

View Answer

Explanation: The volume of the cuboid = lbh

= 8 × 4 × 4

= 128 cm

^{3}

7. What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder?

a) πr^{2}h + 2(2πr^{3})

b) 2πrh – 2(πr^{2})

c) 2πrh + 2(\(\frac {2}{3}\) πr^{2})

d) πr^{2}h – 2(\(\frac {2}{3}\)πr^{3})

View Answer

Explanation: Volume of the article = volume of the cylinder – 2(volume of the hemisphere)

= πr

^{2}h – 2(\(\frac {2}{3}\)πr

^{3})

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)

c) The radius of the cylinder + radius of the cone

d) Height of the cylinder + height of the cone

View Answer

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 cm^{3}

b) 205.6 cm^{3}

c) 168.23 cm^{3}

d) 604 cm^{3}

View Answer

Explanation: Volume of the article = volume of the cylinder – 2(volume of the hemisphere)

= πr

^{2}h – 2(\(\frac {2}{3}\)πr

^{3})

= (3.14 × 5

^{2}× 8) – 2(\(\frac {2}{3}\) × 3.14 × 5

^{3})

= 104.40 cm

^{3}

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

View Answer

Explanation: To find the T.S.A of an article which is made by digging out a hemisphere from each end of a solid cylinder, we need C.S.A of the cylinder and C.S.A of the hemisphere.

T.S.A of the article = C.S.A of the cylinder + 2(C.S.A of the hemisphere).

**Sanfoundry Global Education & Learning Series – Mathematics – Class 10**.

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