Question

A ball of density $\rho$ is dropped from a height $h$ into a container filled with a liquid of density $2\rho$ as shown in the figure. The maximum depth up to which the ball reaches is

• A. $h^{{}^{\prime }}=2h$
• B. $h^{{}^{\prime }}=\frac{h}{2}$
• C. $h^{{}^{\prime }}=3h$
• D. $h^{{}^{\prime }}=h$

Answer 1

The correct option is D $h^{{}^{\prime }}=h$


When the ball is in air, only gravitational force acts on it, but when it enters into the fluid, buoyant force also starts acting on it.
Let the mass of the ball be $m$ and volume be $V$.
Let the maximum depth reached be $h^{{}^{\prime }}$

Gravitational force on the ball $F_G=\rho Vg$

Buoyant force on the ball is $F_B=2\rho Vg$

Applying work-energy theorem, total work done $W_{net}=\Delta K.E$
In this case, $\Delta K.E=0$ [as ball starts from rest and comes to rest at max. depth]
$W_{net}=W_G+W_B$
$=\rho Vg(h+h^{\prime })-2\rho Vg{h}^{\prime }=0$
$\Rightarrow h+h^{{}^{\prime }}=2h^{\prime }$
$\Rightarrow h^{{}^{\prime }}=h$
Thus, option (d) is the correct answer.