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

Ball is dropped from the height of 36 m when the ball will rise at the third bounce
∴ Required height :
$$\eqalign{ & = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times 36 \cr & = \frac{{32}}{3} \cr & = 10\frac{2}{3}\,m \cr} $$

Sum of perimeters of the six faces :
$$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$
Total surface area $$ = 2\left( {lb + bh + lh} \right)$$
$$\eqalign{ & \therefore 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$
Now,
$${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$     $$\left( {lb + bh + lh} \right)$$
$$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 81 - 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 65 \cr} $$
Required length :
$$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$

$$\eqalign{ & 6{a^2} = 150 \cr & \Rightarrow {a^2} = 25 \cr & \Rightarrow a = 5 \cr} $$
∴ Volume :
$${a^3} = {5^3} = 125\,{\text{ c}}{{\text{m}}^3}$$

Let the length of each edge of each cube be a
Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a
Surface area of the cuboid :
$$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$
Sum of surface area of 3 cubes :
$$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$
∴ Required ratio :
$$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$

$$\eqalign{ & \frac{{\pi {r^2}h}}{{2\pi rh}} = \frac{{924}}{{264}} \cr & \Rightarrow r = \left( {\frac{{924}}{{264}} \times 2} \right) \cr & \Rightarrow r = 7\,m \cr} $$
And,
$$\eqalign{ & \therefore 2\pi rh = 264 \cr & \Rightarrow h = \left( {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} \right) \cr & \Rightarrow h = 6\,m \cr} $$
∴ Required ratio :
$$ = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3$$

Let the inner radius of the pipe be r metres
Then,
Volume of water flowing through the pipe in 10 minutes :
$$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times {r^2} \times 7} \right) \times 10} \right]{m^3} \cr & = \left( {220{r^2}} \right){m^3} \cr & \therefore 220{r^2} = 440 \cr & \Rightarrow {r^2} = 2 \cr & \Rightarrow r = \sqrt 2 \, m \cr } $$

Answer :- A right-angled triangle.

Let the heights of two cones be 7x and 3x and their radii be 6y and 7y respectively
Then,
Ratio of volume :
$$\eqalign{ & = \frac{{\frac{1}{3}\pi \times {{\left( {6y} \right)}^2} \times 7x}}{{\frac{1}{3}\pi \times {{\left( {7y} \right)}^2} \times 3x}} \cr & = \frac{{36 \times 7}}{{49 \times 3}} \cr & = \frac{{12}}{7}\,Or\,12:7 \cr} $$

Volume of parallelepiped :
$$\eqalign{ & = \left( {5 \times 3 \times 4} \right)c{m^3} \cr & = 60\,c{m^3} \cr} $$
Volume of cube :
$$\eqalign{ & = {\left( 4 \right)^3}\,c{m^3} \cr & = 64\,c{m^3} \cr} $$
Volume of cylinder :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 3 \times 3 \times 3} \right)c{m^3} \cr & = 84.86\,c{m^3} \cr} $$
Volume of spare :
$$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 3 \times 3 \times 3} \right) \cr & = 113.14\,c{m^3} \cr} $$
So, Option D is correct decreasing order

Volume of new sphere :
$$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( 6 \right)}^3} + \frac{4}{3}\pi \times {{\left( 8 \right)}^3} + \frac{4}{3}\pi \times {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \left\{ {{{\left( 6 \right)}^3} + {{\left( 8 \right)}^3} + {{\left( {10} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3}\pi \times 1728} \right){\text{c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \times {{\left( {12} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr} $$
Let the radius of the new sphere be R
Then,
$$\eqalign{ & \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {\left( {12} \right)^3} \cr & \Rightarrow R = 12\,cm \cr} $$
∴ Diameter :
$$\eqalign{ & = 2R \cr & = 2 \times 12 \cr & = 24\,cm \cr} $$