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

Volume of the tank = 246.4 litres = 246400 cm³
Let the radius of the base be r cm
Then,
$$\eqalign{ & \left( {\frac{{22}}{7} \times {r^2} \times 400} \right) = 246400 \cr & \Rightarrow {r^2} = \left( {\frac{{246400 \times 7}}{{22 \times 400}}} \right) \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14 \cr} $$
∴ Diameter of the base = 2r = 28 cm

Let original height = h and original radius = r
New height = 115% of h $$\frac{23h}{20}$$
New radius = 90% of r $$\frac{9r}{10}$$
Original curved surface area = $$2\pi rh$$
New curved surface area :
$$\eqalign{ & = \left( {2\pi \times \frac{{9r}}{{10}} \times \frac{{23h}}{{20}}} \right) \cr & = \frac{{207}}{{200}} \times 2\pi rh \cr} $$
Increase in curved surface area :
$$\eqalign{ & = \left( {\frac{{207}}{{200}} \times 2\pi rh - 2\pi rh} \right) \cr & = \frac{7}{{200}} \times 2\pi rh \cr} $$
∴ Increase % :
$$\eqalign{ & = \left( {\frac{7}{{200}} \times 2\pi rh \times \frac{1}{{2\pi rh}} \times 100} \right)\% \cr & = 3.5\% \cr} $$

Volume of copper rod :
$$\eqalign{ & = \left( {\pi \times \frac{1}{2} \times \frac{1}{2} \times 8} \right){\text{ c}}{{\text{m}}^3} \cr & = 2\pi \,{\text{ c}}{{\text{m}}^3} \cr} $$
Let the radius of the wire be r cm
Then, volume of wire :
$$\eqalign{ & = \left( \pi {r^2} \times 1800 \right) {\text{cm}}^3 \cr & = 1800\, \pi {r^2} {\text{ cm}}^3 \cr & \therefore 1800\pi {r^2} = 2\pi \cr & \Rightarrow {r^2} = \frac{2}{{1800}} \cr & \Rightarrow {r^2} = \frac{1}{{900}} \cr & \Rightarrow r = \sqrt {\frac{1}{{900}}} = \frac{1}{{30}} \cr} $$

Let the radius of base = r and height = h
Required floor area :
= (4 × 11) m² = 44 m²
Required volume :
= (20 × 11) m³ = 220 m³
So,
$$\eqalign{ & \frac{1}{3}\pi {r^2}h = 220 \cr & \Rightarrow \frac{1}{3} \times 44 \times h = 220 \cr & \Rightarrow h = \frac{{220 \times 3}}{{44}} \cr & \Rightarrow h = 15\,m \cr} $$

$$\eqalign{ & {\text{Volume of cylinder :}} \cr & = \left( {\pi \times 3 \times 3 \times 5} \right)c{m^3} \cr & = 45\pi \,c{m^3} \cr & {\text{Volume of a cone :}} \cr & = \left( {\frac{1}{3}\pi \times \frac{1}{{10}} \times \frac{1}{{10}} \times 1} \right)c{m^3} \cr & = \frac{\pi }{{300}}\,c{m^3} \cr & \therefore {\text{Number of cones :}} \cr & = \left( {45\pi \times \frac{{300}}{\pi }} \right) \cr & = 13500 \cr} $$

Let the radii of the two spheres be r and 4r respectively.
Then,
Required ratio :
$$\eqalign{ & = \frac{{4\pi {r^2}}}{{4\pi {{\left( {4r} \right)}^2}}} \cr & = \frac{{{r^2}}}{{16{r^2}}} \cr & = \frac{1}{{16}} \cr & = 1:16 \cr} $$

Let the edge of the cube be a
Then, volume of the cube = a³
Radius of the sphere $$ = \left( {\frac{a}{2}} \right)$$
Volume of the sphere :
$$\eqalign{ & = \frac{4}{3}\pi {\left( {\frac{a}{2}} \right)^3} \cr & = \frac{{\pi {a^3}}}{6} \cr} $$
∴ Required ratio :
$$\eqalign{ & = {a^3}:\frac{{\pi {a^3}}}{6} \cr & = 6:\pi \cr} $$

Internal radius :
$$\eqalign{ & = \left( {\frac{{11.2}}{2}} \right)cm \cr & = 5.6\,cm \cr} $$
External radius :
$$\eqalign{ & = \left( {5.6 + 0.4} \right)cm \cr & = 6\,cm \cr} $$
Volume of metal :
$$\eqalign{ & = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.6} \right)}^2}} \right\} \times 21} \right]c{m^3} \cr & = \left( {66 \times 11.6 \times 0.4} \right)c{m^3} \cr & = 306.24\,c{m^3} \cr} $$

Volume of sphere :
$$ = \left( {\frac{4}{3}\pi \times 15 \times 15 \times 15} \right)c{m^3}$$
Volume of cone :
$$ = \left( {\frac{1}{3}\pi \times 15 \times 15 \times 15} \right)c{m^3}$$
Volume of wood wasted :
$$\eqalign{ & = \left[ {\left( {\frac{4}{3}\pi \times 15 \times 15 \times 15} \right) - \left( {\frac{1}{3}\pi \times 15 \times 15 \times 15} \right)} \right] \cr & = \left( {\pi \times 15 \times 15 \times 15} \right) \cr} $$
∴ Required percentage :
$$\eqalign{ & = \left( {\frac{{\pi \times 15 \times 15 \times 15}}{{\frac{4}{3}\pi \times 15 \times 15 \times 15}}} \right)\% \cr & = 75\% \cr} $$

Total volume of the body :
= Volume of the cylinder + Volume of the cone + Volume of the hemisphere
$$\eqalign{ & = \pi {r^2}.\,r + \frac{1}{3}\pi {r^2}.\,r + \frac{2}{3}\pi {r^3} \cr & = 2\pi {r^3} \cr & = 2\pi {\left( 2 \right)^3} \cr & = 16\pi \cr} $$