Question

A small ball of density $\rho$ is immersed in a liquid of density $(\sigma >\rho )$ to a depth $h$ and released. The height above the surface of water up to which the ball will jump, is:

• A. $(\frac{\rho }{\sigma }-1)h$
• B. $(\frac{\rho }{\sigma }+1)h$
• C. $(\frac{\sigma }{\rho }-1)h$
• D. $(\frac{\sigma }{\rho }+1)h$

Answer 1

Answer: (C)

$(\frac{\sigma }{\rho }-1)h$

Neglecting the viscous effect of the fluid,
upward force = ma
$\Rightarrow V\sigma g-\rho Vg=\rho Va$
$\Rightarrow \left(\frac{\sigma -\rho }{\rho }\right)g=a$
So, Let velocity of ball after moving $h$ is $V$,
$V^2=U^2+2ah$ as $U=0$
$\Rightarrow V=\sqrt{2(\frac{\sigma }{\rho }-1)gh}$
Now,
After emerging from liquid surface, ball will travel a distance $s$,
$\Rightarrow V_1^2=V^2-2as$, as $V_1=0$
$\Rightarrow 2(\frac{\sigma }{\rho }-1)gh=2\times g\times s$
$\Rightarrow (\frac{\sigma }{\rho }-1)h=s$