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

Let the length of the wire be h
Then,
$$\eqalign{ & \pi \times \frac{3}{{20}} \times \frac{3}{{20}} \times h = \frac{4}{3}\pi \times 4 \times 4 \times 4 \cr & \Rightarrow h = \left( {\frac{{4 \times 4 \times 4 \times 4 \times 20 \times 20}}{{3 \times 3 \times 3}}} \right)cm \cr & \Rightarrow h = \left( {\frac{{102400}}{{27}}} \right)cm \cr & \Rightarrow h = 3792.5\,cm \cr & \Rightarrow h = 37.9\,m \cr} $$

Internal radius, r = 4 cm
External radius, R = 5 cm
Total surface area :
$$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Area of base :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ {\because s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$
∴ Whole surface area of the pyramid :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$

Let,
$$\eqalign{ & OP = OQ = OR = r \cr & \therefore OR = h = r \cr} $$
∴ Curved surface area of the hemisphere = $$2\pi {r^2}$$
Curved surface area of a cone = $$\pi rl$$
Where,
$$\eqalign{ & l = \sqrt {{h^2} + {r^2}} \cr & \,\,\,\,\, = \sqrt {{r^2} + {r^2}} \cr & \,\,\,\,\, = r\sqrt 2 \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{2\pi {r^2}}}{{\pi rl}} \cr & = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{2} \cr & = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1 \cr} $$

Given length of width of swimming pool is 9 m and 12 m respectively
Volume of swimming pool :
$$\eqalign{ & = 9 \times 12 \times \left( {\frac{{1 + 4}}{2}} \right) \cr & = 9 \times 12 \times \frac{5}{2} \cr & = 270\,\text{cu. metre} \cr} $$

Total length of tape required :
= Sum of lengths of edges
= (30 × 4 + 25 × 4 + 20 × 4) cm
= 300 cm

Volume :
$$\eqalign{ & = \left[ {12 \times 9 \times \left( {\frac{{1 + 4}}{2}} \right)} \right]{m^3} \cr & = \left( {12 \times 9 \times 2.5} \right){m^3} \cr & = 270\,{m^3} \cr} $$

Volume of cube = Volume of sheet = (27 × 8 × 1) cm³ = 216 cm³
Edge of cube :
$$\root 3 \of {216} \,cm = 6\,cm$$
Surface area of sheet :
$$\eqalign{ & = 2\left( lb + bh + lh \right) \cr & = 2\left( {27 \times 8 + 8 \times 1 + 27 \times 1} \right){\text{ c}}{{\text{m}}^2} \cr & = \left( {216 + 8 + 27} \right){\text{ c}}{{\text{m}}^2} \cr & = 502{\text{ c}}{{\text{m}}^2} \cr} $$
Surface area of cube :
$$\eqalign{ & = 6{a^2} \cr & = \left( {6 \times {6^2}} \right){\text{ c}}{{\text{m}}^2} \cr & = 216{\text{ c}}{{\text{m}}^2} \cr} $$
∴ Required difference :
$$\eqalign{ & = \left( {502 - 216} \right){\text{ c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Let their edges be a and b
Then,
$$\eqalign{ & \frac{{{a^3}}}{{{b^3}}} = \frac{8}{{27}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \Rightarrow \frac{a}{b} = \frac{2}{3} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{4}{9} \cr & \Rightarrow \frac{{6{a^2}}}{{6{b^2}}} = \frac{4}{9}\,Or\,4:9 \cr} $$

$$\eqalign{ & 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr & \Rightarrow 6r = 4h \cr & \Rightarrow r = \frac{2}{3}h \cr & \Rightarrow r = \left( {\frac{2}{3} \times 6} \right)m \cr & \Rightarrow r = 4\,m \cr} $$