Question

Two solid cones, A and B, are placed in a cylindrical tube as shown in the given figure. The ratio of their capacities is 2:1. Find the volume of the remaining portion of the cylinder. [Assume the thickness of the cones to be negligible.]

• A. $386\text{cubic units}$
• B. $396\text{cubic units}$
• C. $376\text{cubic units}$
• D. $406\text{cubic units}$

Answer 1

The correct option is B $396\text{cubic units}$

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

The ratio of the capacities of cone A and cone B is 2:1.

We observe:
Radius of cone A = Radius of cone B = 3 units

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

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

$\text{Volume of a cone}=\frac{1}{3}\times \pi \times r^2\times h$

So, on putting the values, we get:
$\Rightarrow \frac{\frac{1}{3}\pi \times 3^2\times h_a}{\frac{1}{3}\pi \times 3^2\times h_b}=\frac{2}{1}$

$\Rightarrow \frac{h_a}{h_b}=\frac{2}{1}$

$\Rightarrow h_a=2h_b$

Also, from the figure, we get:
$h_a+h_b=21$
$\Rightarrow 2h_b+h_b=21$
$\Rightarrow 3h_b=21$
$\Rightarrow h_b=7units$

$\therefore h_a=2\times h_b=2\times 7=14units$

So,
Volume of the remaining part of the cylinder
= Volume of the cylinder – Volume of cone A – Volume of cone B
$=\pi r^{2}h-\frac{1}{3}\pi r^2{h}_a-\frac{1}{3}\pi r^2{h}_b$

$=(\pi \times 3^2\times 21)-(\frac{\pi }{3}\times 3^2\times 14)-(\frac{\pi }{3}\times 3^2\times 7)$

$=\pi \times 9\times 21-\frac{\pi }{3}\times 9\times (14+7)$
$=(1-\frac{1}{3})\times \pi \times 9\times 21$

$=\frac{2}{3}\times \frac{22}{7}\times 9\times 21$

$=396\text{cubic units}$