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

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

Let the radius of the cone be r cm
Then,
$$\eqalign{ & \pi \times {\left( 8 \right)^2} \times 2 = \frac{1}{3} \times \pi \times {r^2} \times 6 \cr & \Rightarrow r = 8 \cr} $$
Slant height,
$$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\, = \sqrt {{8^2} + {6^2}} \cr & \,\,\, = \sqrt {100} \cr & \,\,\, = 10\,cm \cr} $$
Curved surface area of cone :
$$\eqalign{ & = \pi rl \cr & = \left( {3.14 \times 8 \times 10} \right){\text{ c}}{{\text{m}}^2} \cr & = 251.2{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = 27 \cr & \Rightarrow r = 81\,cm \cr} $$

$$\eqalign{ & 4\pi {R^2} = 6{a^2} \cr & \Rightarrow \frac{{{R^2}}}{{{a^2}}} = \frac{3}{{2\pi }} \cr & \Rightarrow \frac{R}{a} = \frac{{\sqrt 3 }}{{\sqrt {2\pi } }} \cr} $$
$$\eqalign{ & \therefore \frac{{{\text{Volume of spere}}}}{{{\text{Volume of cube}}}} \cr & = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}} \cr & = \frac{4}{3}\pi {\left( {\frac{R}{a}} \right)^3} \cr & = \frac{4}{3}\pi \frac{{3\sqrt 3 }}{{2\pi \sqrt {2\pi } }} \cr & = \frac{{2\sqrt 3 }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt {12} }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt 6 }}{{\sqrt \pi }} \cr & \text{or, }\sqrt 6 :\sqrt \pi \cr} $$

Volume of sphere :
$$\eqalign{ & = \left( {\frac{4}{3}\pi \times {6^3}} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {288\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$
Volume of each cone :
$$\eqalign{ & = \left( {\frac{1}{3}\pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {12\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$
∴ Number of cone :
$$\eqalign{ & = \frac{{288\pi }}{{12\pi }} \cr & = 24 \cr} $$

Required ratio :
= Volume of cone : Volume of cylinder : Volume of hemisphere
$$\eqalign{ & = \frac{1}{3}\pi {r^2}:\pi {r^2}r:\frac{2}{3}\pi {r^3} \cr & = \frac{1}{3}:1:\frac{2}{3} \cr & = 1:3:2 \cr} $$

Let the height of the cylinder be x cm
Then, radius = (x + 5) m
Curved surface area of the cylinder = $$2\pi rh$$
Now,
$$\eqalign{ & 2\pi \left( {x + 5} \right) \times x = 792 \cr & \Rightarrow 2 \times \frac{{22}}{7} \times \left( {{x^2} + 5x} \right) = 792 \cr & \Rightarrow {x^2} + 5x = \frac{{792 \times 7}}{{44}} = 126 \cr & \Rightarrow {x^2} + 5x - 126 = 0 \cr & \Rightarrow {x^2} + 14x - 9x - 126 = 0 \cr & \Rightarrow x\left( {x + 14} \right) - 9\left( {x + 14} \right) = 0 \cr & \Rightarrow \left( {x - 9} \right)\left( {x + 14} \right) = 0 \cr & \therefore x = 9, - 14{\text{(neglect negative value)}} \cr} $$
∴ Height of cylinder = 9 m
∴ Radius of cylinder = 9 + 5 = 14 m
Volume of cylinder :
$$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 9 \cr & = 5544\,{m^3} \cr} $$

Let the radius of cone and the sphere be R and the height of the cone be H
Volume of sphere $$ = \frac{4}{3}\pi {r^3}$$
Volume of cone $$ = \frac{1}{3}\pi {r^2}h$$
According to given information :
$$\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = 2 \times \frac{1}{3}\pi {R^2}H \cr & \Rightarrow 4R = 2H \cr & \Rightarrow \frac{H}{R} = \frac{4}{2}\,Or\,2:1 \cr} $$

Since the box is an open one, we have :
Internal length = (52 - 4) cm = 48 cm
Internal breadth = (40 - 4) cm = 36 cm
Internal depth = (29 - 2) cm = 27 cm
Volume of the metal used in the box :
= External volume - Internal volume
= [(52 × 40 × 29) - (48 × 36 × 27)] cm³
= (60320 - 46656) cm³
= 13664 cm³
∴ Weight of the box :
$$\eqalign{ & = \left( {\frac{{13664 \times 0.5}}{{1000}}} \right){\text{kg}} \cr & = 6.832\,{\text{kg}} \cr} $$

Clearly, each of the 5 faces of the given cube are glued to a face of another cube
∴ Total surface area of the solid :
= 5 × 5a² + a² = 26a²
= (26 × 3²) cm²
= 234 cm²