Question

A cone is cut into two parts by a horizontal plane passing through the midpoint of its axis. The ratio of the volume of the upper part to the volume of lower part is ............

• A. $1:7$
• B. $1:8$
• C. $7:1$
• D. $7:8$

Answer 1

The correct option is A $1:7$

Let the height of the cone be $H$ and base radius be $R$ units.

Then its volume $V=\frac{1}{3}\pi R²H$ cu. units

As the horizontal plane cuts the cone into two parts through the mid point of its axis, the height of the cone is divided into two equal parts, forming a top cone of height $\frac{H}{2}$ units. If the base radius of the top cone is $r$ units, then $\frac{\frac{r}{H}}{2}=\frac{R}{H}$ [Since the vertical angle of both are same].

Solving, $r=\frac{R}{2}$ units.

Hence volume of the small cone at the top is:

$V=\frac{1}{3}\pi {\left(\frac{R}{2}\right)}^2\left(\frac{H}{2}\right)=\frac{\pi R^{2}H}{24}$ cu. units

So volume of the bottom part (frustum) is:

$V-v=\frac{\pi R^{2}H}{3}-\frac{\pi R^{2}H}{24}=\frac{7}{24}\pi R^{2}H$ cu. units

Therefore, ratio of their volumes, Top part : Bottom part that is:

$\frac{\pi R^{2}H}{24}:\frac{7}{24}\pi R^{2}H=1:7$

Hence, the ratio of the volume of the upper part to the volume of lower part is $1:7$.