Question

A hemisphere and a solid cone have a common base. The centre of mass of the common structure coincides with the centre of the common base. If R is the radius of hemisphere and h is the height of the cone, then :

• A. $\frac{h}{R}=\sqrt{3}$
• B. $\frac{h}{R}=\frac{1}{\sqrt{3}}$
• C. $\frac{h}{R}=3$
• D. $\frac{h}{R}=\frac{1}{3}$

Answer 1

The correct option is A $\frac{h}{R}=\sqrt{3}$

Consider the common center of hemisphere and circular base of cone as origin.

Since, the shape is symmetrical in x direction, $x_{cm}=0$

$y_{cm}=\frac{M_{cone}{y}_{cone}+M_{hs}{y}_{hs}}{M_{cone}+M_{hs}}=0$

$\Rightarrow M_{cone}{y}_{cone}+M_{hs}{y}_{hs}=0$

Let density of both the shapes be $\rho$

$\therefore M_{cone}=\frac{1}{3}\pi R^{2}h\rho$

Similarly, $M_{hs}=\frac{2}{3}\pi R^3\rho$

Also, $y_{cone}=\frac{h}{4}$

and, $y_{hs}=-\frac{3R}{8}$

$\Rightarrow \frac{1}{3}\pi R^{2}h\rho \times \frac{h}{4}+\frac{2}{3}\pi R^3\rho \times (-\frac{3R}{8})=0$

$\Rightarrow \frac{h^2}{R^2}=3$

$\Rightarrow \frac{h}{R}=\sqrt{3}$

Correct option is A