Dynamics and Viscous Forces on a Table Tennis Ball Pushed Underwater

To determine which option(s) is(are) correct, we analyze each statement step-by-step using fundamental principles of fluid mechanics and dynamics.

1. Given Data

• 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, $g = 10 \text{ m s}^{-2}$
• Value of $\pi = \frac{22}{7}$
• Density of water, $\rho_w = 1 \times 10^3 \text{ kg m}^{-3}$
• Viscosity of water, $\eta = 1 \times 10^{-3} \text{ Pa}\cdot\text{s}$

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2. Buoyant Force and Weight of the Ball

The volume of the spherical ball is: $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \times \left(\frac{22}{7}\right) \times \left(\frac{3}{2} \times 10^{-2}\right)^3 = \frac{99}{7} \times 10^{-6} \text{ m}^3$

The buoyant force ($F_B$) acting vertically upwards on the completely submerged ball is: $F_B = \rho_w V g = (10^3) \times \left(\frac{99}{7} \times 10^{-6}\right) \times 10 = \frac{99}{7} \times 10^{-2} \text{ N} = 0.1414 \text{ N}$

The gravitational force ($mg$) acting vertically downwards is: $mg = \left(\frac{22}{7} \times 10^{-3}\right) \times 10 = \frac{22}{7} \times 10^{-2} \text{ N}$

The net upward force excluding viscous force is: $F_{\text{net, ex}} = F_B - mg = \frac{99 - 22}{7} \times 10^{-2} = \frac{77}{7} \times 10^{-2} = 0.11 \text{ N}$

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3. Analysis of Option A

To slowly push the ball down to depth $d$, an external force $F_{\text{ext}}$ equal to the net upward buoyant force must be applied downwards: $F_{\text{ext}} = F_{\text{net, ex}} = 0.11 \text{ N}$

The work done in slowly pushing the ball down to a depth $d = 0.7 \text{ m}$ is: $W_{\text{ext}} = F_{\text{ext}} \times d = 0.11 \text{ N} \times 0.7 \text{ m} = 0.077 \text{ J}$

Thus, Option A is correct.

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4. Analysis of Option B

If viscous forces are neglected, the net constant acceleration $a$ of the ball upwards is: $a = \frac{F_{\text{net, ex}}}{m} = \frac{0.11}{\frac{22}{7} \times 10^{-3}} = \frac{0.11 \times 7}{0.022} = 35 \text{ m s}^{-2}$

Using the equation of motion $v^2 = u^2 + 2ad$ with $u = 0$: $v^2 = 2 \times 35 \times 0.7 = 49 \implies v = 7 \text{ m/s}$

Thus, Option B is correct.

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5. Analysis of Option C

After emerging from the water surface with speed $v = 7 \text{ m/s}$, the ball rises freely in the air under gravity to a maximum height $H$: $H = \frac{v^2}{2g} = \frac{49}{2 \times 10} = 2.45 \text{ m}$

Thus, Option C is incorrect (since it states $H = 1.4 \text{ m}$).

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6. Analysis of Option D

The viscous force in water is given by Stokes' Law: $F_v = 6 \pi \eta R v$

The maximum viscous force occurs at the maximum speed reached by the ball inside water, which occurs just as it reaches the surface ($v_{\max} = 7 \text{ m/s}$): $F_{v, \text{max}} = 6 \times \left(\frac{22}{7}\right) \times (1 \times 10^{-3}) \times \left(\frac{3}{2} \times 10^{-2}\right) \times 7 = 1.98 \times 10^{-3} \text{ N}$

The ratio of the magnitude of the net force excluding viscous force to the maximum viscous force is: $\text{Ratio} = \frac{F_{\text{net, ex}}}{F_{v, \text{max}}} = \frac{0.11}{1.98 \times 10^{-3}} = \frac{110 \times 10^{-3}}{1.98 \times 10^{-3}} = \frac{110}{1.98} = \frac{500}{9}$

Thus, Option D is correct.

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Conclusion

The correct options are A, B, and D.