Question

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

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

Answer 1

The Correct Option is A, B, D

Work Done, Speed, and Viscous Force Calculation 

The work done is:

\[ W = (\text{Buoyancy force} - \text{Weight}) \cdot d = \rho g \cdot \frac{4}{3} \pi r^3 - mg \cdot d \] Substituting: \[ W = \frac{4}{3} \cdot \pi \cdot \left( 3 \times 10^{-2} \right)^3 \cdot 10 \cdot 0.7 \cdot 1000 - \frac{3}{4} = 0.077 \, \text{J} \]

For Speed:

\[ \frac{1}{2} m v^2 = W \quad \Rightarrow \quad v = \sqrt{\frac{2W}{m}} = 7 \, \text{m/s} \] Also, the viscous force is maximum when \( v = 7 \, \text{m/s} \), hence: \[ (F_v)_{\text{max}} = 6 \pi \eta r v = 6 \times 22 \times 7 \times 10^{-3} \times 3 \times 10^{-2} = 18 \times 11 \times 10^{-5} \, \text{N} \]

Now:

\[ \frac{F_{\text{net}}}{(F_v)_{\text{max}}} = \frac{500}{9} \]

Correct Options:

Options (1) and (2), (3) are correct.