Question

A small ball of mass $m$ is released from rest from a height $h$ above the liquid surface. The density of ball is $\frac{\rho }{3}:$
(assume no splash)

• A. if $h\ge H$, then the ball will be able to enter the liquid of higher density
• B. if $h\ge \frac{7H}{2}$, then the ball will be able to strike the bottom of the vessel
• C. if $h=2H$, then the ball will cross the interface of two liquids with kinetic energy = $mgH$
• D. if $h=0$ then ball will float on the surface

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A if $h=2H$, then the ball will cross the interface of two liquids with kinetic energy = $mgH$
B if $h=0$ then ball will float on the surface
C if $h\ge H$, then the ball will be able to enter the liquid of higher density
D if $h\ge \frac{7H}{2}$, then the ball will be able to strike the bottom of the vessel

(D):
If $h=0$, the ball will float in the liquid since ${\rho }_b=\frac{1}{3}{\rho }_l$
(A):
The velocity of the ball when it hits the liquid surface after dropping from a height h is $v_b=\sqrt{2gh}$
Two forces act on the ball when it is inside the liquid: weight of the ball and upward thrust due to buoyancy.
$f_b=m_b{a}_b=f_{thrust}-w_b=v_b{\rho }_{l}g-v_b\frac{{\rho }_l}{3}g=\frac{2}{3}{v}_b{\rho }_{l}g$
$\therefore v_b\frac{{\rho }_l}{3}{a}_b=\frac{2}{3}{v}_b{\rho }_{l}g$
$a_b=2g$ upward
If the ball has to enter the liquid of density $2\rho$, its velocity at the boundary of the liquids having densities $\rho$ and $2\rho$
will be non zero.
Applying one dimensional eqn
$v^2-u^2=2as$
$v^2-(\sqrt{2gh}{)}^2=2(-2g)\frac{H}{2}$ We have taken $a_b=-2g$ since it is deceleration.
$\therefore v^2=2gh-2gH=2g(h-H)$
Since $v^2$ > 0 at boundary of the two liquids, $h\ge H$.
(B):
When the ball enters the liquid of density $2\rho$, the upward thrust will increase,
$f_b=m_b{a}_b=f_{thrust}-w_b=v_{b}2{\rho }_{l}g-v_b\frac{{\rho }_l}{3}g=\frac{5}{3}{v}_b{\rho }_{l}g$
$\therefore v_b\frac{{\rho }_l}{3}{a}_b=\frac{5}{3}{v}_b{\rho }_{l}g$
$a_b=5g$ upward
If the balls drops from a height h above the liquid surface, the velocity at which the ball strikes the water surface is $u=\sqrt{2gh}$.
Velocity at the boundary of both the liquids is
$(v_b^{\prime }{)}^2-u^2=-2\times 2g\times \frac{H}{2}=2gh-2\times 2g\times \frac{H}{2}$
Putting $h=\frac{7H}{2}$
$(v_b^{\prime }{)}^2=2g\frac{7H}{2}-2gH=5gH$
Again, applying one dimensional equation of motion to the layer having density $2\rho$,
$(v_b^{\prime \prime }{)}^2-5gH=-2(5g)\frac{H}{2}$ where $v_b^{\prime \prime }$ is the velocity of the liquid at the bottom of the container.
$(v_b^{\prime \prime }{)}^2=5gH-2(5g)\frac{H}{2}=0$
$\therefore v^2=2gh-2gH=2g(h-H)$
Thus, the ball will be able to just reach the bottom of the container if the velocity is $\frac{7H}{2}$
(C):
If $h=2H$,
then, the ball strikes the liquid surface with velocity $\sqrt{2g\times 2H}=\sqrt{4gH}$
Velocity at the boundary of two liquids is
$v^2=(\sqrt{4gH}{)}^2-2\times 2g\times \frac{H}{2}=2gH$
$\therefore \frac{1}{2}{m}_b{v}^2=m_{b}gH=mgH$