Question

A ball of density ${\rho }_b$ falls from rest from a point P on to the surface of a liquid of density ${\rho }_l$ in time T. It falls, enters the liquid, stops, moves up and returns to P in a total time 3T. Neglect viscosity, surface tension and splashing. The ratio $\frac{{\rho }_l}{{\rho }_b}$ will be equal to:

• A. 1
• B. 3
• C. 2
• D. 9

Answer 1

The correct option is B 3

It is given that:

The ball returns to the point $P$ in total time $3T$.
Time to fall to the surface of the liquid is $T$.

If the ball finally reaches point $P$ and comes to rest, then time taken to reach the point $P$ from the surface of the liquid is also $T$.
Hence $3T-2T=T$ is the time spent inside the liquid.

Lets assume acceleration inside the liquid is $a_b$ and volume of the ball is $V_b$.

Force on the ball $=$ gravitation force of ball $-$ upward thrust due to buoyancy

$\therefore V_b{\rho }_b{a}_b=V_b{\rho }_{l}g-V_b{\rho }_{b}g$

Acceleration inside the liquid is $a_b=\frac{V_b{\rho }_{l}g-V_b{\rho }_{b}g}{V_b{\rho }_b}=g(\frac{{\rho }_l}{{\rho }_b}-1)$ ................(1)

It has been found out that time spend inside the liquid is T, hence time to reach the bottom of the liquid container will be $\frac{T}{2}$. The ball when touched the surface of liquid, it has acquired a velocity of $gT$ when dropped from the rest.

Let decelaration in liquid is $a_b$, then using equation of motion

$v=u-a_b\frac{T}{2}$, (As ball stops hence $v=0$ and $u=gT$)

$\therefore a_b=\frac{gT}{\frac{T}{2}}=2g$,
Substituting $a_b=2g$ in eqn (1)
$2g=g(\frac{{\rho }_l}{{\rho }_b}-1)$
$\Rightarrow (\frac{{\rho }_l}{{\rho }_b}-1)=2$
$\Rightarrow \frac{{\rho }_l}{{\rho }_b}=3$