Question

A uniform solid cone of height $H$ is cut at a distance $\times $ below its vertex, parallel to its base and the upper part is discarded. The $C.M.$ of the remaining part is found to be $50\%$ lower, w.r.t. to the base. The ratio, $r=\frac{x}{H}$ satisfies.

• A. $7r^{3}-8r^{4}=1$
• B. $r=\frac{1}{2^{1 / 3}}$
• C. $r=\frac{1}{2}$
• D. $7r^{3}-6r^{4}=1$

Answer 1

Answer: (D)

The final $C.M.$ is at $\frac{H}{8}$ above the base. The volume, and consequently the mass of a cone (of fixed vertical angle) is proportional to the cube of its height. Taking the origin at apex. $\frac{7 H}{8}=\frac{H^{3} \left(\frac{3 H}{4}\right) - x^{3} \left(\frac{3 x}{4}\right)}{H^{3} - x^{3}}$
Rearranging, we get the required condition.
$\Rightarrow \frac{7 H}{8}\times \frac{4}{3}=\frac{H^{4} - x^{4}}{H^{3} - x^{3}}$
$\Rightarrow \frac{7 H}{6}=\frac{H^{4} - x^{4}}{H^{3} - x^{3}}$
$\Rightarrow \frac{7}{6}=\frac{1 - r^{4}}{1 - r^{3}}$
$\Rightarrow 7\left(1 - r^{3}\right)=6\left(1 - r^{4}\right)$
$\Rightarrow 7-7r^{3}=6-6r^{4}$
$\Rightarrow 7r^{3}-6r^{4}=1$