A ball whose density is 0.4 × 10³ kg/m³ falls into water from a height of 9 cm . To what depth does the ball sink

The correct option is 6 cm.

To find the depth to which the ball sinks in the water, we can break the motion of the ball into two parts: free fall in the air and retarded motion inside the water.

Step 1: Motion in Air (Free Fall)
The ball is dropped from a height $h=9\text{cm}$ above the water surface. Let $g$ be the acceleration due to gravity. The velocity $v$ of the ball just as it reaches the water surface is given by the equation of motion:
$v^2=2gh$

Step 2: Motion in Water (Retardation)
Once the ball is completely submerged in water, it experiences two opposing forces:
1. A downward gravitational force: $F_g=V{\rho }_{b}g$
2. An upward buoyant force: $F_b=V{\rho }_{w}g$
where $V$ is the volume of the ball, ${\rho }_b=0.4\times {10}^3{\text{kg/m}}^3$ is the density of the ball, and ${\rho }_w=1.0\times {10}^3{\text{kg/m}}^3$ is the density of water.

Since the density of water is greater than the density of the ball, the buoyant force is greater than the gravitational force, resulting in a net upward force. This net upward force acts as a retarding force, slowing the ball down. The net upward force is:
$F_{\text{net}}=F_b-F_g=Vg({\rho }_w-{\rho }_b)$

The acceleration (retardation) $a$ of the ball is:
$a=\frac{F_{\text{net}}}{m}=\frac{Vg({\rho }_w-{\rho }_b)}{V{\rho }_b}=g\left(\frac{{\rho }_w}{{\rho }_b}-1\right)$

Substituting the given values of densities into this expression:
$a=g\left(\frac{1.0\times {10}^3}{0.4\times {10}^3}-1\right)=g(2.5-1)=1.5g$

Step 3: Calculating the Depth
Let $d$ be the depth to which the ball sinks before its velocity becomes zero. Using the equation of motion for a retarding acceleration $a$:
$v_f^2=v^2-2ad$

Since the final velocity at the lowest point is $v_f=0$, we have:
$0=2gh-2ad$
$ad=gh$
$d=\frac{gh}{a}$

Substitute the value of $a=1.5g$ and $h=9\text{cm}$:
$d=\frac{g\times 9}{1.5g}=\frac{9}{1.5}=6\text{cm}$

Therefore, the ball sinks to a depth of 6 cm in the water.