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

External radius = 3 cm
Internal radius = (3 - 0.5) cm = 2.5 cm
Volume of the metal :
$$\eqalign{ & = \left[ {\frac{4}{3} \times \frac{{22}}{7} \times \left\{ {{{\left( 3 \right)}^3} - {{\left( {2.5} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{91}}{8}} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{{143}}{3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 47\frac{2}{3}{\text{ c}}{{\text{m}}^3} \cr} $$

Let the height of the cylinder be h cm
Then,
$$\eqalign{ & \frac{4}{3}\pi \left[ {{{\left( 4 \right)}^3} - {{\left( 2 \right)}^3}} \right] = \pi \times {4^2} \times h \cr & \Rightarrow \frac{4}{3} \times \pi \times 56 = \pi \times 16h \cr & \Rightarrow h = \frac{{4 \times 56}}{{3 \times 16}} \cr & \Rightarrow h = \frac{{14}}{3}\,cm \cr} $$

Let the radius of the hemisphere be be r cm
Then, radius of the cylinder = r cm
Height of the cylinder = r cm
∴ Required ratio :
$$\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr} $$

The volume of cone should be maximum
∴ Radius of the base of cone :
$$\eqalign{ & = \frac{{{\text{Edge of cube}}}}{2} \cr & = \frac{{4.2}}{2} \cr & = 2.1\,{\text{dm.}} \cr} $$
Height of cone = Edge of cube = 4.2 dm.
∴ Volume of cone :
$$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times 2.1 \times 2.1 \times 4.2} \right){\text{cu}}{\text{.dm}}{\text{.}} \cr & = 19.404\,{\text{cu}}{\text{.dm}}{\text{.}} \cr} $$

Volume of hemisphere :
$$\eqalign{ & = \left( {\frac{2}{3}\pi \times 3 \times 3 \times 3} \right){m^3} \cr & = \left( {18\pi } \right){m^3} \cr} $$

Required length :
$$\eqalign{ & = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr & = \sqrt {225} \,m \cr & = 15\,m \cr} $$

Let the radius of the cylinder be r and height be h
Then, r + h = 19
Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$
Now,
$$\eqalign{ & 2\pi r\left( {h + r} \right) = 1672 \cr & \Rightarrow 2\pi r \times 19 = 1672 \cr & \Rightarrow 38\pi r = 1672 \cr & \therefore \pi r = \frac{{1672}}{{38}} = 44\,m \cr & \therefore r = \frac{{44 \times 7}}{{22}} = 14\,m \cr} $$
∴ Height = 19 - 14 = 5 m
Volume of cylinder :
$$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 5 \cr & = 22 \times 2 \times 14 \times 5 \cr & = 3080\,{m^3} \cr} $$

Let length = x cm
Then,
$$\eqalign{ & x \times 28 \times 5 \times \frac{{25}}{{1000}} = 112 \cr & \Rightarrow x = \left( {112 \times \frac{{1000}}{{25}} \times \frac{1}{{28}} \times \frac{1}{5}} \right) \cr & \Rightarrow x = 32\,cm \cr} $$

Clearly,
$$l$$ = (20 - 4) cm = 16 cm
b = (15 - 4) cm = 11 cm and
h = 2 cm
∴ Outer surface area of the box :
$$\eqalign{ & = \left[ {2\left( {l + b} \right) \times h} \right] + lb \cr & = \left[ {\left\{ {2\left( {16 + 11} \right) \times 2} \right\} + 16 \times 11} \right] \cr & = \left( {108 + 176} \right) \cr & = 284{\text{ c}}{{\text{m}}^2} \cr} $$

Clearly, we have :
V₁ = x³,
V₂ = (2x)³ = 8x³
V₃ = (3x)³ = 27x³
V₄ = (4x)³ = 64x³
(i) V₁ + V₂ + 2V₃
= x³ + 8x³ + 2 × 27x³
= 63x³ < V₄

(ii) V₁ + 4V₂ + V₃
= x³ + 4 × 8x³ + 27 x³
= 60x³ < V₄

(iii) 2(V₁ + V₃) + V₂
= 2 (x³ + 27x³) + 8x³
= 64x³ = V₄