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

Let the radius of the sphere and that of the right circular cylinder be r
Then,
Volume of the cylinder $$ = \pi {r^2}h$$
Volume of the sphere $$ = \frac{4}{3}\pi {r^3}$$
$$\eqalign{ & \therefore \pi {r^2}h = \frac{4}{3}\pi {r^3} \cr & \Rightarrow 3h = 4r \cr & \Rightarrow 3h = 2d \cr & \Rightarrow \frac{h}{2} = \frac{d}{3} \cr} $$

Volume of hemisphere = $$\frac{2}{3}\pi {r^3}$$
Volume of biggest sphere :
= Volume of sphere with diameter r
$$\eqalign{ & = \frac{4}{3}\pi {\left( {\frac{r}{2}} \right)^3} \cr & = \frac{1}{6}\pi {r^3} \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{\frac{2}{3}\pi {r^3}}}{{\frac{1}{6}\pi {r^3}}} \cr & = \frac{4}{1}i.e.,4:1 \cr} $$

Volume of earth dug out :
= (240 × 180 × 0.25) m³
= 10800 m³
Let the depth of the drain-let be h metres
Then, Volume of earth dug out :
= [{(260 × 200) - (240 × 180)}h ] m³
= (8800h) m³
∴ 8800h = 10800
⇒ h = $$\frac{10800}{8800}$$
⇒ h = $$\frac{27}{22}$$
⇒ h = 1.227 m

Let lb = 2x, bh = 3x and lh = 4x
Then,
$$\eqalign{ & 24{x^3} = {\left( {lbh} \right)^2} = 9000 \times 9000 \cr & \Rightarrow {x^3} = 375 \times 9000 \cr & \Rightarrow x = 150 \cr} $$
So, lb = 300, bh = 450, lh = 600 and lbh = 9000
$$\eqalign{ & \therefore h = \frac{{9000}}{{300}} = 30 \cr & l = \frac{{9000}}{{450}} = 20\& \cr & b = \frac{{9000}}{{600}} = 15 \cr} $$
Hence, shortest side = 15 cm

Number of small cube formed :
$$\eqalign{ & = \left( {\frac{{4 \times 4 \times 4}}{{1 \times 1 \times 1}}} \right) \cr & = 64 \cr} $$
Total surface area of the small cubes :
$$\eqalign{ & = \left[ {64 \times \left( {6 \times {1^2}} \right)} \right]{\text{c}}{{\text{m}}^2} \cr & = 384{\text{ c}}{{\text{m}}^2} \cr} $$

Curved surface area :
$$\eqalign{ & = 2\pi rh \cr & = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 22.5} \right){{\text{m}}^2} \cr & = 495{\text{ }}{{\text{m}}^2} \cr} $$
∴ Cost of plastering :
= Rs. (495 × 3)
= Rs. 1485

Let the drop in the water level be h cm
Then,
$$\eqalign{ & \Rightarrow \frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times h = 11000 \cr & \Rightarrow h = \left( {\frac{{11000 \times 7 \times 4}}{{22 \times 35 \times 35}}} \right){\text{ cm}} \cr & \Rightarrow h = \frac{{80}}{7}{\text{ cm}} \cr & \Rightarrow h = 11\frac{3}{7}{\text{ cm}} \cr} $$

Let the length of each side of the square base be x metres
Then,
$$\eqalign{ & {x^2} + {x^2} = {\left( 3 \right)^2} \cr & \Rightarrow 2{x^2} = 9 \cr & \Rightarrow {x^2} = \frac{9}{2} \cr & \Rightarrow x = \frac{3}{{\sqrt 2 }} \cr} $$
∴ Volume of parallelopiped :
$$\eqalign{ & = \left( {\frac{3}{{\sqrt 2 }} \times \frac{3}{{\sqrt 2 }} \times 10} \right){{\text{m}}^{\text{3}}} \cr & = \frac{{90}}{2}{{\text{m}}^{\text{3}}} \cr & = 45\,{{\text{m}}^{\text{3}}} \cr} $$

Let the radius and height of the cone be 3x and 4x respectively
Then,
$$\eqalign{ & \frac{1}{3} \times \frac{{22}}{7} \times {\left( {3x} \right)^2} \times 4x = \frac{{2112}}{7} \cr & \Rightarrow \frac{{264}}{7}{x^3} = \frac{{2112}}{7} \cr & \Rightarrow {x^3} = \frac{{2112}}{{264}} \cr & \Rightarrow {x^3} = 8 \cr & \Rightarrow x = 2 \cr} $$
∴ Radius = 6 cm, Height = 8 cm
Slant height :
$$\eqalign{ & = \sqrt {{6^2} + {8^2}} \,cm \cr & = \sqrt {100} \,cm \cr & = 10\,cm \cr} $$

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
$$\eqalign{ & = 2\pi rH + \pi rl \cr & = \left[ {\frac{{22}}{7}\left( {2 \times 12 \times 11 + 12 \times 13} \right)} \right]{{\text{m}}^2} \cr & = \left[ {\frac{{22}}{7}\left( {264 + 156} \right)} \right]{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 420} \right){{\text{m}}^2} \cr & = 1320\,{{\text{m}}^2} \cr} $$