Volume And Surface Area - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Clearly}}, \cr & {\text{We have }}r = 3\,{\text{cm}}\,{\text{and}}\,h = 4\,{\text{cm}} \cr & \therefore \,\,\,{\text{Volume}} \cr & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^3} \cr & = 12\pi \,{\text{c}}{{\text{m}}^3} \cr} $$

$$\eqalign{ & {\text{1}}\,{\text{hectare}} = 10000\,{m^2} \cr & {\text{So,}}\,{\text{Area}} = \left( {1.5 \times 10000} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15000\,{m^2} \cr & {\text{Depth}} = \frac{5}{{100}}m = \frac{1}{{20}}m \cr & \therefore {\text{Volume}} = \left( {{\text{Area}}\, \times \,{\text{Depth}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {15000 \times \frac{1}{{20}}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 750\,{m^3} \cr} $$

$$\eqalign{ & 2\left( {15 + 12} \right) \times h = 2\left( {15 \times 12} \right) \cr & \Rightarrow h = \frac{{180}}{{27}}m = \frac{{20}}{3}m \cr & \therefore {\text{Volume}} = \left( {15 \times 12 \times \frac{{20}}{3}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200\,{m^3} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{wire}}\,{\text{be h}} \cr & {\text{Radius}} = \frac{1}{2}mm = \frac{1}{{20}}cm.\,{\text{Then}}, \cr & \Rightarrow \frac{{22}}{7} \times \frac{1}{{20}} \times \frac{1}{{20}} \times h = 66 \cr & \Rightarrow h = {\frac{{66 \times 20 \times 20 \times 7}}{{22}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8400\,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\,meters \cr} $$

$$\eqalign{ & {\text{External}}\,{\text{radius}} = 4\,cm \cr & {\text{Internal}}\,{\text{radius}} = 3\,cm \cr & {\text{Volume}}\,{\text{of}}\,{\text{iron}} \cr & = \left( {\frac{{22}}{7} \times \left[ {{{\left( 4 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times 21} \right)c{m^3} \cr & = \left( {\frac{{22}}{7} \times 7 \times 1 \times 21} \right)c{m^3} \cr & = 462\,c{m^3} \cr & \therefore {\text{Weight}}\,{\text{of}}\,{\text{iron}} = \left( {462 \times 8} \right)gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3696\,gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.696\,kg \cr} $$

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.

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

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

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

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