Practice MCQ Questions and Answer on Volume and Surface Area

11.

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. The area of the canvas required for the tent is :

• (A) 1300 m2
• (B) 1310 m2
• (C) 1320 m2
• (D) 1330 m2

Show Answer

 Radius, r = 12 m Height of conical part, h = (16 - 11) m = 5 m Slant height of conical part, $$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\, = \sqrt {{{\left( {12} \right)}^2} + {{\left( 5 \right)}^2}} \cr & \,\,\,\, = \sqrt {169} \cr & \,\,\,\, = 13\,m \cr} $$ Height of cylindrical part, H = 11 m Area of canvas required : = Curved surface area of cylinder + Curved surface area of cone$$\eqalign{ & = 2\pi rH + \pi rl \cr & = \left[ {\frac{{22}}{7}\left( {2 \times 12 \times 11 + 12 \times 13} \right)} \right]{{\text{m}}^2} \cr & = \left[ {\frac{{22}}{7}\left( {264 + 156} \right)} \right]{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 420} \right){{\text{m}}^2} \cr & = 1320\,{{\text{m}}^2} \cr} $$

12.

How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?

• (A) 5600
• (B) 6000
• (C) 6400
• (D) 7200

Show Answer

 $$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{bricks}} \cr & = \frac{{{\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{wall}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{1}}\,{\text{brick}}}} \cr & = {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \cr & = 6400 \cr} $$

13.

The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is :

• (A) 1 : 1
• (B) 1 : 2
• (C) 2 : 1
• (D) 2 : 3

Show Answer

 Let the radius of the sphere be r Then, radius of the cylinder = r Height of the cylinder = 2r Surface area of sphere = $$4\pi {{\text{r}}^2}$$ Surface area of the cylinder = $$2\pi {\text{r}}(2r) = 4\pi {{\text{r}}^2}$$ ∴ Required ratio : = $$4\pi {{\text{r}}^2}$$ : $$4\pi {{\text{r}}^2}$$ = 1 : 1

14.

A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?

• (A) 2 : 1
• (B) 3 : 2
• (C) 25 : 18
• (D) 27 : 20

Show Answer

 $$\eqalign{ & {\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}} \cr & = \left( {{3^3} + {4^3} + {5^3}} \right) = 216\,c{m^3} \cr & {\text{Let}}\,{\text{the}}\,{\text{edge}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}}\,{\text{be}}\,a \cr & So,\,{a^3} = 216\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 6\,cm \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{{6 \times \left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}}} \cr & = \frac{{50}}{{36}} \cr & = 25:18 \cr} $$

15.

For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?

• (A) 24%
• (B) 26.5%
• (C) 30%
• (D) 45%

Show Answer

 Volume of the sphere : $$ = \left[ {\frac{4}{3}\pi {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3}$$ Surface area of the sphere : $$ = \left[ {4\pi {{\left( {10} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2}$$ ∴ Required percentage : $$\eqalign{ & = \left[ {\frac{{4\pi {{\left( {10} \right)}^2}}}{{\frac{4}{3}\pi {{\left( {10} \right)}^3}}}} \times 100 \right]\% \cr & = 30\% \cr} $$

16.

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m³, then the rise in the water level in the tank will be:

• (A) 20 cm
• (B) 25 cm
• (C) 35 cm
• (D) 50 cm

Show Answer

 $$\eqalign{ & {\text{Total}}\,{\text{Volume}}\,{\text{of}}\,{\text{water}}\,{\text{displaced}} \cr & = \left( {4 \times 50} \right){m^3} = 200\,{m^3} \cr & \therefore {\text{Rise}}\,{\text{in}}\,{\text{water}}\,{\text{level}} \cr & = \left( {\frac{{200}}{{40 \times 20}}} \right)m \cr & = 0.25\,m \cr & = 25\,cm \cr} $$

17.

Find the number of coins 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

• (A) 430
• (B) 440
• (C) 450
• (D) 460

Show Answer

 Volume one coin : $$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{75}}{{100}} \times \frac{{75}}{{100}} \times \frac{2}{{10}}} \right){\text{c}}{{\text{m}}^3} \cr & = \frac{{99}}{{280}}{\text{ c}}{{\text{m}}^3} \cr} $$ Volume of larger cylinder : $$ = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10} \right){\text{ c}}{{\text{m}}^3}$$ Number of coins : $$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \times \frac{{280}}{{99}}} \right) \cr & = 450 \cr} $$

18.

What length of solid cylinder 2 cm in diameter must be taken to cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick and 15 cm long ?

• (A) 42.3215 cm
• (B) 44.0123 cm
• (C) 44.0625 cm
• (D) 44.6023 cm

Show Answer

 External radius = 6 cm Internal radius = (6 - 0.25) = 5.75 cm Volume of material in hollow cylinder : $$\eqalign{ & = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.75} \right)}^2}} \right\} \times 15} \right]{\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times 11.75 \times 0.25 \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times \frac{{1175}}{{100}} \times \frac{{25}}{{100}} \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{11 \times 705}}{{56}}} \right){\text{c}}{{\text{m}}^3} \cr} $$ Let the length of solid cylinder be h Then, $$\eqalign{ & \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr & \Rightarrow h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr & \Rightarrow h = 44.0625\,{\text{cm}} \cr} $$

19.

The volume of a sphere is $$2145\frac{11}{21}\text{c}{\text{m}}^3.$$   Its radius is equal to :

• (A) 7 cm
• (B) 8 cm
• (C) 9 cm
• (D) None of these

Show Answer

 $$\eqalign{ & \frac{4}{3} \times \frac{{22}}{7} \times {r^3} = \frac{{45056}}{{21}} \cr & \Rightarrow {r^3} = \left( {\frac{{45056}}{{21}} \times \frac{3}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^3} = 512 \cr & \Rightarrow r = \root 3 \of {512} \cr & \Rightarrow r = 8\,cm \cr} $$

20.

A metallic cone of radius 12 cm and height 24 cm is ,melted and made into spheres of radius 2 cm each. How many spheres are there ?

• (A) 108
• (B) 120
• (C) 144
• (D) 180

Show Answer

 $$\eqalign{ & {\text{Number of spheres :}} \cr & = \frac{{{\text{Volume of cone}}}}{{{\text{Volume of 1 sphere}}}} \cr & = \frac{{\frac{1}{3}\pi \times 12 \times 12 \times 24}}{{\frac{4}{3}\pi \times 2 \times 2 \times 2}} \cr & = 108 \cr} $$