Question

Problem - 11.8: A ball of relative density 0.8 falls into
water from a height of 2m. Find the depth to which the
ball will sink (neglect viscous forces)​

Answer 1

When the ball is dropped from the height, it has only potential energy due to gravity.

So the total initial energy of the ball,

$\sf{\longrightarrow U_1=mgh }$

Taking $\sf{m=\rho V}$ where $\rho$ is density and $\sf{V}$ is volume of the ball.

$\sf{\longrightarrow U_1=\rho Vgh}$

At the depth $\sf{h',}$ the ball experiences force of gravity downwards and buoyant force upwards.

So the total final energy of the ball will be,

$\sf{\longrightarrow U_2=mgh'-F_bh'}$

If $\sf{\rho_w}$ is density of water,

$\sf{\longrightarrow U_2=\rho Vgh'-\rho_w Vgh' }$

because buoyant force is weight of displaced liquid.

$\sf{\longrightarrow U_2=(\rho-\rho_w)Vgh' }$

By energy conservation,

$\sf{\longrightarrow U_1=U_2}$

$\sf{\longrightarrow\rho Vgh=(\rho-\rho_w)Vgh' }$

$\sf{\longrightarrow h'=\dfrac{\rho h}{\rho-\rho_w}}$

$\sf{\longrightarrow h'=\dfrac{\dfrac{\rho}{\rho_w}\cdot h}{\dfrac{\rho}{\rho_w}-1}}$

Taking $\sf{\dfrac{\rho}{\rho_w}=s,}$ the relative density of the body,

$\sf{\longrightarrow\underline{\underline{h'=\dfrac{sh}{s-1}}}}$

In this question,

• $\sf{s=0.8}$

• $\sf{h=2\ m}$

Then,

$\sf{\longrightarrow h'=\dfrac{sh}{s-1}}$

$\sf{\longrightarrow h'=\dfrac{0.8\times2}{0.8-1}}$

$\sf{\longrightarrow\underline{\underline{h'=-8\ m}}}$

Negative sign depicts the downward motion of the body through the liquid.