A table tennis ball has radius and mass It is slowly pushed down into a swimming pool to a depth of below the water surface and then released from rest. It emerges from the water surface at speed , without getting wet, and rises up to a height . Which

Question

A table tennis ball has radius $(\frac{3}{2}) \times 10^{- 2} m$ and mass $(\frac{22}{7}) \times 10^{- 3} kg .$ 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: $π = \frac{22}{7}, g = 10$ ${m s}^{- 2}$ , density of water = $1 \times 10^{3} {kg m}^{- 3}$ , viscosity of water = $1 \times 10^{- 3} Pa-s .$ ] [2024]

Smart Achievers · Accepted Answer

(1, 2, 4)

$(1)$ $Work done in pushing the ball to the depth d = 0.7 m$

$W_{all} = k_{f} - k_{i} = 0$

$w_{g} + w_{B} + w_{v} + w_{ext} = 0$

$m g d - ρ_{w} \cdot v \cdot g d - 6 π η r v d + w_{ext} = 0$

$w_{ext} = ρ_{w} \cdot v \cdot g d - m g d$

$= (1000 \times \frac{4}{3} \times \frac{22}{7} \times {(\frac{3}{2} \times 10^{- 2})}^{3} - \frac{22}{7} \times 10^{- 3}) g d$

$w_{ext} = \frac{22}{7} \times 10^{- 3} [\frac{9}{2} - 1] \times 10 \times 0.7 = \frac{22}{7} \times 10^{- 3} \times \frac{7}{2} \times 7$

$w_{ext} = 77 \times 10^{- 3} J = 0.077 J$

$so option (1) is correct.$

$(2) When ball is released at bottom same work i.e. 0.077 J$ $is done on ball$

$∴$ $\frac{1}{2} m v^{2} = 0.077$ $∴$ $v = \sqrt{\frac{2 \times 0.077}{\frac{22}{7} \times 10^{- 3}}} = 7 m/s$

$So option (2) is correct$

$(3) Height H = \frac{v^{2}}{2 g} = \frac{49}{20} = 2.45 m$

$so option (3) is incorrect$

$Net force F_{net} = v d g - v ρ g = 0.11 N$

$And viscous force is maximum when v = 7 m/s$

$∴$ ${(F_{v})}_{\max} = 6 π η r v = 6 \times \frac{22}{7} \times 10^{- 3} \times (\frac{3}{2} \times 10^{- 2}) \times 7$

$= 18 \times 11 \times 10^{- 5} N$

$∴$ $\frac{F_{net}}{{(F_{v})}_{\max}} = \frac{0.11}{18 \times 11 \times 10^{- 5}} = \frac{500}{9}$

$so, option (4) is correct$

Solution

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 $v = 7 m/s$ .

3 If we neglect the viscous force in water, then the height $H = 1.4 m$ .

4 The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $\frac{500}{9} .$

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Solution

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.

3 If we neglect the viscous force in water, then the height H=1.4 m.

4 The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 5009.

Want Us to Email you About Special Offers & Updates?

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 $v = 7 m/s$ .

3 If we neglect the viscous force in water, then the height $H = 1.4 m$ .

4 The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $\frac{500}{9} .$