Question

A small ball of mass $m$ is released from rest from a height $h$ above the liquid surface. The density of the ball is $\frac{\rho }{3}$. Then which one of the following is incorrect? (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 $h=H$, then the ball will cross the interface of two liquids with $kineticenergy=4mgH$
• D. If $h=0$ then the ball will float on the surface

Answer 1

The correct option is C

If $h=H$, then the ball will cross the interface of two liquids with $kineticenergy=4mgH$

Step 1: Given data

Mass of the ball$\to m$

The height it's dropped from$\to h$

The density of the ball$\to \frac{\rho }{3}$

Step 2: Formula & Concept

1. In this particular question, we need to use the concept of conservation of energy
2. The potential energy the ball lost is converted into work done by the buoyance force by the liquid on the ball
3. Work done by the buoyance force is $v\rho gh$
4. Where $v$ is the volume, $\rho$is the density of the liquid, $g$ is the acceleration due to gravity, $h$ is the distance the ball travels in liquid.
5. The volume of the ball $v=3m/\rho$

The explanation for the correct answer:

In case of option C:

If $h=H$, Kinetic energy with which the ball crosses the interface of two liquids will be potential energy lost - energy lost due to buoyance force

⇒$KE=mg[H+\frac{H}{2}]-\frac{3m}{\rho }\rho g\frac{H}{2}$

⇒ $KE=\frac{3}{2}mg-\frac{3}{2}mg$

⇒ $KE=0$

The explanation for the incorrect answer:

In case of option A:

For the ball to enter the liquid of higher density : $mg[h+H/2]\ge v\rho gH/2$

⇒ $mg[h+H/2]\ge \frac{3}{2}mgH$

⇒ $h\ge H/2$

In case of option B:

For the ball to strike the bottom:$mg[h+H]\ge \frac{3m}{\rho }g\frac{H}{2}(\rho +2\rho )$

⇒ $mg[h+H]\ge \frac{9}{2}mgH$

⇒ $h\ge \frac{7}{2}H$

In case of option D:

1. It is given that $h=0$
2. The body won't have any energy at the time of contact with water and we know that the density of the liquid is higher than the density of the ball.
3. The ball will just float on the surface of the liquid without sinking.

If $h=H$, then the ball will cross the interface of two liquids with $kineticenergy=4mgH$

Hence, Option C is incorrect.