Question

A table tennis ball has radius \( \frac{3}{2} \times 10^{-2} \, \text{m} \) and mass \( \frac{22}{7} \times 10^{-3} \, \text{kg} \). It is slowly pushed down into a swimming pool to a depth \( d = 0.7 \, \text{m} \) below the water surface and then released from rest. It emerges from the water surface at speed \( v \), without getting wet, and rises to a height \( H \). Which of the following option(s) is(are) correct?\(\text{(Given: } \pi = \frac{22}{7}, g = 10 \, \text{m/s}^2, \text{density of water} = 1 \times 10^3 \, \text{kg/m}^3, \text{viscosity of water} = 1 \times 10^{-3} \, \text{Pa.s).}\)

• A. The work done in pushing the ball to the depth \( d \) is \( 0.077 \, \text{J}. \)
• B. If we neglect the viscous force in water, then the speed \( v = 7 \, \text{m/s}. \)
• C. If we neglect the viscous force in water, then the height \( H = 1.4 \, \text{m}. \)
• D. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is \( \frac{500}{9}. \)

Hint:

For work done in fluid mechanics, consider buoyancy and weight forces.

Answer 1

The Correct Option is A

The work done: \[ W = \text{(Buoyancy force - Weight)} \cdot d = \left(\rho g \frac{4}{3} \pi r^3 - mg\right) \cdot d. \] Substituting: \[ W = \frac{4}{3} \cdot \pi \cdot \left(\frac{3}{2} \times 10^{-2}\right)^3 \cdot 10 \cdot 0.7 \cdot \left(1000 - \frac{3}{4}\right) = 0.077 \, \text{J}. \] For speed: \[ \frac{1}{2} mv^2 = W \quad \Rightarrow \quad v = \sqrt{\frac{2W}{m}} = 7 \, \text{m/s}. \] \begin{align*} \text{Also, viscous force is maximum when } v &= 7 \, \text{m/s}
\therefore (F_v)_{\text{max}} &= 6\pi\eta rv
&= 6 \times \frac{22}{7} \times 10^{-3} \left(\frac{3}{2} \times 10^{-2}\right) \times 7
&= 18 \times 11 \times 10^{-5} \, \text{N}
\text{Now,} \quad \frac{F_{\text{net}}}{(F_v)_{\text{max}}} &= \frac{500}{9}
\end{align*} Thus, Options (1) and (2) ,(3) are correct.