Question

2 solid cones X and Y are placed in a cylindrical tube as shown below. The ratio of their capacities is 3:1. Find the volume of the remaining portion of the cylinder.

• A. $\frac{128}{3}{units}^3$
• B. $\frac{108}{3}\pi {units}^3$
• C. $\frac{128}{3}\pi {units}^3$
• D. $128{units}^3$

Answer 1

The correct option is C $\frac{128}{3}\pi {units}^3$

The diameter of the cylinder is 4 units.
Thus, the radius of the cylinder is 2 units.

The capacities of X and Y are in the ratio 3:1

Radius of cone X = Radius of cone Y = 2 unit

Let the heights of cone X and Y be $h_x$ unit and $h_y$ unit respectively.

Assuming the thickness to be negligible, we can say that the volumes of cones X and Y are in the ratio 3:1

$\Rightarrow \frac{\frac{1}{3}\times \pi \times 3^2\times h_x}{\frac{1}{3}\times \pi \times 3^2\times h_y}=\frac{3}{1}$

$\Rightarrow \frac{h_x}{h_y}=\frac{3}{1}$

$\Rightarrow h_x=3h_y$

The total height of cylinder = 16 units.

So,
$h_x+h_y=16$

$\Rightarrow 3h_y+h_y=16$

$\Rightarrow 4h_y=16$

$\Rightarrow h_y=4units$

$\text{So,}{h}_x=3\times h_y=3\times 4=12units$

∴ Volume of the remaining part of the cylinder = Volume of the cylinder - Volume of Cone A - Volume of Cone B

$=\pi r^{2}h-\frac{1}{3}\pi r^2{h}_x-\frac{1}{3}\pi r^2{h}_y$

$=(\pi \times 2^2\times 16)-(\frac{\pi }{3}\times 2^2\times 12)-(\frac{\pi }{3}\times 2^2\times 4)$

$=\pi \times 4\times 16-\frac{\pi }{3}\times 4\times (12+4)$

$=(1-\frac{1}{3})\times \pi \times 4\times 16$

$=\frac{2}{3}\times \pi \times 4\times 16$

$=\frac{128}{3}{units}^3$

So, volume of remaining part of cylinder is $\frac{128}{3}{units}^3.$