Question

Two solid cones A and B are placed in a cylindrical tube as shown in the figure below. The ratio of their capacities is 2:1. Find the volume of the remaining portion of the cylinder.

• A. $406{cm}^3$
• B. $396{cm}^3$
• C. $376{cm}^3$
• D. $386{cm}^3$

Answer 1

The correct option is B

$396{cm}^3$

The diameter of the cylinder is 6 cm.
Thus, the radius of the cylinder is 3 cm.

The capacities of A and B are in the ratio 2:1

Radius of cone A
= Radius of cone B = 3 cm

Let the heights of cone A and B be $h_a$ and $h_b$ respectively.

Assuming the thickness to be negligible, we can say that the volumes of cones A and B are in the ratio 2:1

$\Rightarrow \frac{\frac{1}{3}\pi \times 3^2\times h_a}{\frac{1}{3}\pi \times 3^2\times h_b}=\frac{2}{1}$
$\Rightarrow \frac{h_a}{h_b}=\frac{2}{1}$
$\Rightarrow h_a=2h_b$

Also,
$h_a+h_b=21$
$\Rightarrow 2h_b+h_b=21$
$\Rightarrow 3h_b=21$
$\Rightarrow h_b=7cm$
$h_a=2\times h_b=2\times 7=14cm$

Volume of the remaining part of the cylinder
= Volume of the cylinder - Volume of cone A - Volume of cone B
$=\pi r^{2}h-\frac{1}{3}\pi r^2{h}_a-\frac{1}{3}\pi r^2{h}_b$
$=(\pi \times 3^2\times 21)-(\frac{\pi }{3}\times 3^2\times 14)-(\frac{\pi }{3}\times 3^2\times 7)$
$=\pi \times 9\times 21-\frac{\pi }{3}\times 9\times (14+7)$
$=(1-\frac{1}{3})\times \pi \times 9\times 21$
$=\frac{2}{3}\times \frac{22}{7}\times 9\times 21$
$=396{cm}^3$