Question

A ball whose density is $0.4\times {10}^{3}kg/m^3$ falls into water from a height of $9cm$. To what depth does the ball sink.

• A. $9cm$
• B. $6cm$
• C. $4.5cm$
• D. $2.25cm$

Answer 1

Answer: (B)

The ball is dropped from a height of $h_0=9cm$. The velocity of the ball just before entering the water surface will be $v_0=\sqrt{2g{h}_0}$, where $h_0=9cm=0.09m$

After entering into the water with velocity $v_0$, the buoyant force will work on it in opposite direction, and will stop it's motion. The net force on ball under water will be $F_{net}=\text{Buoyant force - weight of ball}=V\times \rho \times g-V\times \sigma \times g$

$\Rightarrow F_{net}=Vg(\rho -\sigma )$

Where $\rho$ and $\sigma$ are the densities of water and ball respectively.

$F_{net}$ force will work in the opposite direction of motion of the ball and will give retardation of ${}^{\prime }{a}^{\prime }$

Where $a=\frac{F_{net}}{m}=$ $\left(\frac{F_{net}}{V\sigma }\right)=\frac{Vg(\rho -\sigma )}{V\sigma }=(\frac{\rho }{\sigma }-1)g$

By putting the density of water is $\rho ={10}^{3}kg/m^3$ and the density if ball is $\sigma =0.4\times {10}^{3}kg/m^3$ in above equation we get,

$\Rightarrow a=1.5g$

By using a third equation of motion we can write $v^2=u^2-2ah$, where ${}^{\prime }{v}^{\prime }$ is the final velocity of ball under water, ${}^{\prime }{u}^{\prime }$ is initial velocity of ball under water and ${}^{\prime }{h}^{\prime }$ is the depth covered by ball in the water, ${}^{\prime }{a}^{\prime }$ is the retardation due to net force working in opposite direction.

Due to retardation, the ball will stop eventually at some depth ${}^{\prime }{h}^{\prime }$, hence the final velocity of the ball under water $v=0$, and initial velocity of a ball under water is same as $v_0$, so $u=v_0$

Hence we can write $h=\frac{(v_0{)}^2}{2a}$, $a=1.5g$ and $v_0=\sqrt{2g{h}_0}$

By putting all the given values in the above equation we get,

$\Rightarrow h=\frac{2h_0}{3}$

$\Rightarrow h=\frac{2\times 9cm}{3}=6cm$

Hence Correct answer is $B$.