Question

A small ball of density ${\rho }_0$ is released from rest from the surface of a liquid whose density varies with depth $h$ as $\rho =\frac{{\rho }_0}{2}(a+\beta h)$Mass of the ball is $m$.Select the most appropriate option:

• A. The particle will execute SHM.
• B. The maximum speed of the ball is $\frac{2-a}{\sqrt{2\beta }}$.
• C. Both (a) and (b) are correct.
• D. Both (a) and (b) are wrong.

Answer 1

The correct option is A The particle will execute SHM.

$\rho =\frac{{\rho }_o}{2}(a+\beta h)$

$V=\frac{m}{{\rho }_o}$

Downward force, $F=mg-\frac{m}{{\rho }_o}\frac{{\rho }_o}{2}(a+\beta h)g$

$F=mg((1-\frac{a}{2})-\frac{\beta h}{2})$

So F is proportional to displacement from center position which is $mg(1-\frac{a}{2})$ below surface

Therefore, the motion is SHM.

A = $mg(1-\frac{a}{2})$

$k=\frac{mg\beta }{2}$

Also, $k=m{\omega }^2$

=> $\omega =\sqrt{\frac{g\beta }{2}}$

$v_{max}=\omega A=\frac{(2-a)\sqrt{g}}{\sqrt{2}\beta }$

Answer A