## Volume and Surface Area Questions and Answers Part-8

1. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:
a) 12 kg
b) 60 kg
c) 72 kg
d) 96 kg

Explanation: Volume of water displaced
= (3 x 2 x 0.01) m3
= 0.06 m3.
Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.

2. 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

Explanation:
\eqalign{ & {\text{Total}}\,{\text{Volume}}\,{\text{of}}\,{\text{water}}\,{\text{displaced}} \cr & = \left( {4 \times 50} \right){m^3} = 200\,{m^3} \cr & {\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}

3. The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.
a) 30π m2
b) 40π m2
c) 60π m2
d) 80π m2

Explanation:
\eqalign{ & l = 10\,m \cr & h = 8\,m \cr & So,\,r = \sqrt {{l^2} - {h^2}} = \sqrt {{{\left( {10} \right)}^2} - {8^2}} = 6\,m \cr & {\text{Curved}}\,{\text{surface}}\,{\text{area}} \cr & \pi \,rl = \left( {\pi \times 6 \times 10} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60\pi \,{m^2} \cr}

4. A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
a) 49 m2
b) 50 m2
c) 53.5 m2
d) 55 m2

Explanation:
\eqalign{ & {\text{Area}}\,{\text{of}}\,{\text{the}}\,{\text{wet}}\,{\text{surface}} \cr & = \left[ {2\left( {lb + bh + lh} \right) - lb} \right] \cr & = 2\left( {bh + lh} \right) + lb \cr & = \left[ {2\left( {4 \times 1.25 + 6 \times 1.25} \right) + 6 \times 4} \right]{m^2} \cr & = 49\,{m^2} \cr}

5. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:
a) 4830
b) 5120
c) 6420
d) 8960

Explanation: Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box
= (32 x 20 x 8) m3
= 5120 m3

6. If the volume and surface area of a sphere are numerically the same, then its radius is :
a) 1 unit
b) 2 units
c) 3 units
d) 4 units

Explanation:
\eqalign{ & \frac{4}{3}\pi {r^3} = 4\pi {r^2} \cr & r = 3 \text{ units} \cr}

7. A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is :
a) 2.5 cm
b) 3 cm
c) 3.5 cm
d) 4 cm

Explanation: Let the radius of the sphere be r cm
\eqalign{ & \frac{4}{3}\pi {r^3} = \pi \times {\left( {0.1} \right)^2} \times 3600 \cr & {r^3} = 36 \times \frac{3}{4} \cr & {r^3} = 27 \cr & r = 3\,cm \cr}

8.The capacities of two hemispherical vessels are 6.4 litres and 21.6 litres. The areas of inner curved surfaces of the vessels will be in the ratio of :
a) $$\sqrt 2$$ : $$\sqrt 3$$
b) 2 : 3
c) 4 : 9
d) 16 : 81

Explanation: Let their radii be R and r
\eqalign{ & \frac{{\frac{2}{3}\pi {R^3}}}{{\frac{2}{3}\pi {r^3}}} = \frac{{6.4}}{{21.6}} \cr & {\left( {\frac{R}{r}} \right)^3} = \frac{8}{{27}} \cr & {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \frac{R}{r} = \frac{2}{3} \cr}
Ratio of curved surface area :
$$= \frac{{2\pi {R^2}}}{{2\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} = \frac{4}{9}\,or\,4:9$$

9. The base of a pyramid is an equilateral triangle of side 1 m. If the height of the pyramid is 4 metres, then the volume is :
a) 0.550 m3
b) 0.577 m3
c) 0.678 m3
d) 0.750 m3

Explanation: Area of the base :
\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr}
Volume of pyramid :
\eqalign{ & = \left( {\frac{1}{3} \times \frac{{\sqrt 3 }}{4} \times 4} \right){m^3} \cr & = \left( {\frac{{\sqrt 3 }}{3}} \right){m^3} \cr & = \left( {\frac{{1.732}}{3}} \right){m^3} \cr & = 0.577\,{m^3} \cr}.

10.Each side of a cube is decreased by 25%. Find the ratio of the volume of the original cube and the resulting cube = ?
a) 64 : 1
b) 27 : 64
c) 64 : 27
d) 8 : 1

\eqalign{ & = \frac{{1000}}{{421.875}} \cr & = \frac{{1000000}}{{421875}} \cr & = \frac{{64}}{{27}}\,Or\,64:27 \cr}