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

Given radii of three metallic spheres be $${r_1},\,{r_2},\,{r_3}$$   are 6 cm, 8 cm and 10 cm respectively
Let the radius of the new sphere be R
$$\eqalign{ & \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \left( {r_1^3 + r_2^3 + r_3^3} \right) \cr & \Rightarrow \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \left( {{6^3} + {8^3} + {{10}^3}} \right) \cr & \Rightarrow {R^3} = \left( {216 + 512 + 1000} \right) \cr & \Rightarrow {R^3} = 1728 \cr & \Rightarrow R = \root 3 \of {12 \times 12 \times 12} \cr & \Rightarrow R = 12 \cr} $$
∴ Diameter = 24 cm

Volume of earth dug out :
$$\eqalign{ & = \left( {5 \times 4.5 \times 2.1} \right){{\text{m}}^{\text{3}}} \cr & = 47.25\,{{\text{m}}^{\text{3}}} \cr} $$
Area over which earth is spread :
$$\eqalign{ & = \left( {13.5 \times 2.5 - 5 \times 4.5} \right){{\text{m}}^2} \cr & = \left( {33.75 - 22.5} \right){{\text{m}}^2} \cr & = 11.25\,{{\text{m}}^2} \cr} $$
$$\eqalign{ & \therefore {\text{Rise in level}} = \frac{{{\text{Volume}}}}{{{\text{Area}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{47.25}}{{11.25}}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.2\,m \cr} $$

Let, length = l, breadth = b, height = h
Then,
x = lb, y = bh, z = lh
Let, V be the volume of the cuboid
Then, V = lbh
$$\eqalign{ & \therefore xyz = lb \times bh \times lh \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {lbh} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {V^2} \cr & \,\,\,\,\,\,\,\,\,\,\,or\,V = \sqrt {xyz} \cr} $$

Number of cubes :
$$\eqalign{ & = \left( {\frac{{100 \times 100 \times 100}}{{10 \times 10 \times 10}}} \right) \cr & = 1000 \cr} $$

Clearly, the cylinder formed by rolling the paper along its length has height 18 cm and circumference of base 30 cm i.e.,
$$\eqalign{ & h = 18{\text{ cm and }} \cr & {\text{2}}\pi r = 30 \cr & Or,r = \frac{{30}}{2} \times \frac{7}{{22}} \cr & Or,r = \frac{{105}}{{22}} \cr} $$
∴ Volume :
$$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}} \times 18} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{14175}}{{11}}{\text{ c}}{{\text{m}}^3} \cr} $$
The cylinder formed by rolling the paper along its breadth has height 30 cm and circumference of base 18 cm i.e.,
$$\eqalign{ & h = 30{\text{ cm and }} \cr & {\text{2}}\pi r = 18 \cr & Or,r = \frac{{18}}{2} \times \frac{7}{{22}} \cr & Or,r = \frac{{63}}{{22}} \cr} $$
∴ Volume :
$$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times \frac{{63}}{{22}} \times \frac{{63}}{{22}} \times 30} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{8505}}{{11}}{\text{ c}}{{\text{m}}^3} \cr} $$
Required ratio :
$$\eqalign{ & = \frac{{14175}}{{11}}:\frac{{8505}}{{11}} \cr & = 5:3 \cr} $$

Let original radius = r and original height = h
Original volume = $$\pi {r^2}h$$
New radius = 125% of r = $$\frac{5r}{4}$$
Let new height = H
Then,
$$\eqalign{ & \pi {r^2}h = \pi {\left( {\frac{{5r}}{4}} \right)^2} \times H \cr & Or,\,H = \frac{{16}}{{25}}h \cr} $$
Decrease in height :
$$\eqalign{ & = \left( {h - \frac{{16h}}{{25}}} \right) \cr & = \frac{{9h}}{{25}} \cr} $$
∴ Decrease % :
$$\eqalign{ & = \left( {\frac{{9h}}{{25}} \times \frac{1}{h} \times 100} \right)\% \cr & = 36\% \cr} $$

Volume of water flown into the tank :
$$\eqalign{ & = \left( {50 \times 44 \times 0.07} \right){{\text{m}}^3} \cr & = 154\,{{\text{m}}^3} \cr} $$
Volume of water flowing through the pipe in 1 hour :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 0.07 \times 0.07 \times 5000} \right){{\text{m}}^3} \cr & = 77\,{\text{m}} \cr} $$
∴ Required time $$ = \left( {\frac{{154}}{{77}}} \right) = 2{\text{ hours}}$$

Volume flown in conical vessel :
$$\eqalign{ & = \frac{1}{2}\pi \times {\left( {20} \right)^2} \times 24 \cr & = 3200\pi \cr} $$
Volume flown in 1 minute :
$$\eqalign{ & = \left( {\pi \times \frac{{2.5}}{{10}} \times \frac{{2.5}}{{10}} \times 1000} \right) \cr & = 62.5\pi \cr} $$
∴ Time taken :
$$\eqalign{ & = \left( {\frac{{3200\pi }}{{62.5\pi }}} \right) \cr & = 51\operatorname{minutes} 12\operatorname{seconds} \cr} $$

Let the original radius be r
Then, original volume = $$\frac{4}{3}\pi {r^3}$$
New area = 2r
∴ New volume :
$$\eqalign{ & = \frac{4}{3}\pi {\left( {2r} \right)^3} \cr & = 8 \times \frac{4}{3}\pi {r^3} \cr & = 8 \times {\text{Original volume}} \cr} $$

Let their radii be R and r
Then,
$$\eqalign{ & \frac{{\frac{4}{3}\pi {R^3}}}{{\frac{4}{3}\pi {r^3}}} = \frac{{64}}{{27}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^3} = \frac{{64}}{{27}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{4}{3}} \right)^3} \cr & \Rightarrow \frac{R}{r} = \frac{4}{3} \cr} $$
Ratio of surface areas :
$$\eqalign{ & = \frac{{4\pi {R^2}}}{{4\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{4}{3}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{9} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16:9 \cr} $$