Question

A small ball of relative density 0.8m falls into water from a height of 2m. The depth to which the ball will sink, is(neglect viscous forces)
a) 8m
b) 2cm
c) 6cm
d) 4cm

Answer 1

Hint: The relative density of the substance is defined as the ratio of density of the substance to that of density of water. The ball falls from a height of 2m and hence it will enter the water with some initial velocity. As the ball progresses inside the water, it will experience the buoyant force such that its velocity will become zero at an instant of time. Hence we can determine the depth to which the ball will sink using Newton’s kinematic equations.

Formula used:
${{v}^{2}}-{{u}^{2}}=2as$
${{F}_{B}}={{\rho }_{w}}gV$
${{F}_{G}}=mg$

Complete step-by-step answer:
Initially the body approaches the water from a height of s=2m from rest i.e. u=0m/s. Since the body is accelerating due to gravity ‘g’, the velocity with which it will just enter the water is equal to,
$\begin{align}
  & {{v}^{2}}-{{u}^{2}}=2as \\
 & {{v}^{2}}-{{(0)}^{2}}=2g2m \\
 & \Rightarrow {{v}^{2}}=4g \\
 & \therefore v=\sqrt{4g}m/s \\
\end{align}$
Let us say a body of volume ‘V’, mass ‘m’ and density ${{\rho }_{B}}$ is immersed fully in water where in the density of water is ${{\rho }_{w}}$ . Than the buoyant force on the body in the upward direction is given by,
${{F}_{B}}={{\rho }_{w}}gV$
As the body is sinking there is downward force due to gravity on the body as well. Hence the net force (F)on the body is,
$\begin{align}
  & F={{F}_{G}}-{{F}_{B}} \\
 & \Rightarrow ma=mg-{{\rho }_{w}}gV \\
 & \Rightarrow a=g-\dfrac{{{\rho }_{w}}g}{(m/V)}, \\
 & \because {{\rho }_{B}}=m/V \\
 & \Rightarrow a=g-\dfrac{g}{({{\rho }_{B}}/{{\rho }_{w}})} \\
 & \because {{\rho }_{B}}/{{\rho }_{w}}=0.8 \\
 & \Rightarrow a=g-\dfrac{g}{0.8}=-\dfrac{0.2}{0.8}g \\
 & \therefore a=-\dfrac{1}{4}g \\
\end{align}$
The ball enters the water with the speed of $\sqrt{4g}m/s$and after some time it will come to rest due to negative acceleration ‘a’. Hence from Newton’s kinematic equation, the depth ‘h’ to which the ball sinks is,
$\begin{align}
  & {{v}^{2}}-{{u}^{2}}=2ah \\
 & \Rightarrow {{(0)}^{2}}-{{\left( \sqrt{4g} \right)}^{2}}=2\left( -\dfrac{1}{4} \right)gh \\
 & \Rightarrow 4=\dfrac{1}{2}h \\
 & \therefore h=8m \\
\end{align}$
Therefore the ball will sink to a depth of 8m from the surface.

Note: In the question it is mentioned that the viscous force of the water is to be ignored. Hence the above equation of motion of the ball in water is valid. The density of any substance is defined as the mass per unit volume of that substance.