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Question

Physics Question on Fluid Mechanics

A table tennis ball has radius (3/2)×102(3/2) × 10^{−2} m and mass (22/7)×103(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: π=22/7,g=10ms2\pi = 22/7, g = 10 m s ^{−2} , density of water = 1×103kg m31 × 10^3 kg\ m^{−3} , viscosity of water = 1×1031 × 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

The work done in pushing the ball to the depth 𝑑 is 0.077 J

Explanation

Solution

The correct option is (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. and (D): The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.