$\text{A table tennis ball has radius }\left(\frac{3}{2}\right)\times {10}^{-2}m\text{and mass}\left(\frac{22}{7}\right)\times {10}^{-3}\mathrm{kg}.\text{It is slowly pushed down into a}$

$\text{swimming pool to a depth of}\mathrm{ d}=0.7m\text{below the water surface and then released from rest.}\text{It emerges}$

$\text{from the water surface at speed}\mathrm{ v},\text{without getting wet, and rises up to a height}\mathrm{ H}.\text{Which of the following}$

$\text{option(s) is(are) correct?}$
$[\text{Given:}\pi =\frac{22}{7},g=10\frac{m}{s^2},\text{density of water}=1\times {10}^3\frac{\mathrm{kg}}{m^3},\text{viscosity of water}=1\times {10}^{-3}\mathrm{Pa}.s]$

The correct options are:

• The work done in pushing the ball to the depth d is 0.077J .
• If we neglect the viscous force in water, then the speed v = 7m/s .
• The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

Step 1: Given Data and Preliminary Calculations

Radius of the ball, $r=\frac{3}{2}\times {10}^{-2}\text{ m}$

Mass of the ball, $m=\frac{22}{7}\times {10}^{-3}\text{ kg}$

Depth, $d=0.7\text{ m}$

Acceleration due to gravity, $g=10\text{ m/s}{}^2$

Density of water, ${\rho }_w={10}^3\text{ kg/m}{}^3$

Viscosity of water, $\eta ={10}^{-3}\text{ Pa}·\text{s}$

Value of $\pi =\frac{22}{7}$

First, let us calculate the volume of the table tennis ball, $V$:

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times {\left(\frac{3}{2}\times {10}^{-2}\right)}^3\text{ m}{}^3$

$V=\frac{4}{3}\times \frac{22}{7}\times \frac{27}{8}\times {10}^{-6}=\frac{99}{7}\times {10}^{-6}\text{ m}{}^3$

Now, calculate the buoyant force ($F_B$):

$F_B=V{\rho }_{w}g=\left(\frac{99}{7}\times {10}^{-6}\right)\times {10}^3\times 10=\frac{99}{7}\times {10}^{-2}\text{ N}=0.1414\text{ N}$

Calculate the gravitational force ($W$):

$W=mg=\left(\frac{22}{7}\times {10}^{-3}\right)\times 10=\frac{22}{7}\times {10}^{-2}\text{ N}=0.0314\text{ N}$

Step 2: Calculate the work done in pushing the ball to depth $d$

The net upward force excluding viscous force is:

$F_{\text{net}}=F_B-mg=\frac{99}{7}\times {10}^{-2}-\frac{22}{7}\times {10}^{-2}=\frac{77}{7}\times {10}^{-2}=11\times {10}^{-2}\text{ N}=0.11\text{ N}$

To slowly push the ball down, an external force equal to $F_{\text{net}}$

Work done, $W_{\text{ext}}=F_{\text{net}}\times d=0.11\times 0.7=0.077\text{ J}$.

Thus, the first option is correct.

Step 3: Calculate the speed $v$ neglecting viscous force

Using the work-energy theorem, work done by the net force during the release of the ball over depth $d$ equals the gain in kinetic energy at the surface:

$F_{\text{net}}\times d=\frac{1}{2}m{v}^2$

$0.077=\frac{1}{2}\times \left(\frac{22}{7}\times {10}^{-3}\right)\times v^2$

$\frac{77}{1000}=\frac{11}{7000}{v}^2$

$v^2=\frac{77\times 7000}{1000\times 11}=7\times 7=49$

$v=7\text{ m/s}$

Thus, the second option is correct.

Step 4: Ratio of net force (excluding viscous force) to maximum viscous force

The maximum speed attained by the ball in water when neglecting viscous force is at the surface, $v_{\text{max}}=7\text{ m/s}$.

The maximum viscous force according to Stokes' law is:

$F_{\text{viscous, max}}=6\pi \eta r{v}_{\text{max}}$

$F_{\text{viscous, max}}=6\times \left(\frac{22}{7}\right)\times {10}^{-3}\times \left(\frac{3}{2}\times {10}^{-2}\right)\times 7=6\times 22\times \frac{3}{2}\times {10}^{-5}=198\times {10}^{-5}\text{ N}$

The ratio of $F_{\text{net}}$ to $F_{\text{viscous, max}}$ is:

$\text{Ratio}=\frac{11\times {10}^{-2}}{198\times {10}^{-5}}=\frac{11\times {10}^3}{198}=\frac{11000}{198}=\frac{500}{9}$

Thus, the fourth option is also correct.