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: = 22/7, g = 10 m s-2, density of water = 1 103 kg m-3, viscosity of water = 1 10-3 Pa-s.]
- The work done in pushing the ball to the depth d is 0.077 J.
- If we neglect the viscous force in water, then the speed v = 7 m/s.
- If we neglect the viscous force in water, then the height H = 1.4 m.
- 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 work done in pushing the ball to the depth d is 0.077 J., (2) If we neglect the viscous force in water, then the speed v = 7 m/s., (4) The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.
Solution
V = 43 r3 = 43 227 3.375 10-6 = 997 10-6 m3
Buoyancy = Vg = 0.1414 N, weight = mg = 0.0314 N; net upward force (excluding viscous) F = 0.11 N.
(A) Slow push: W = Fd = 0.11 0.7 = 0.077 J — correct.
(B) No viscosity: 12mv2 = 0.077 v2 = 0.154(22/7) 10-3 = 49, v = 7 m/s — correct.
(C) H = v22g = 4920 = 2.45 m — wrong.
(D) Maximum viscous force (at v = 7 m/s): 6 rv = 6 227 10-3 1.5 10-2 7 = 1.98 10-3 N; ratio = 0.111.98 10-3 = 5009 — correct.
Answer: (A), (B), (D)