JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Fluid Mechanics

A table tennis ball has radius

(3/2) \times 1 0^{- 2} m

and mass

(22/7) \times 1 0^{- 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 = 10 ms^{- 2}

, density of water

= 1 \times 1 0^{3} kg m^{- 3}

, viscosity of water

= 1 \times 1 0^{- 3} Pa

-s.]

(A)The work done in pushing the ball to the depth $d$ is 0.077 J.✓ Correct
(B)If we neglect the viscous force in water, then the speed $v = 7 m / s$ .✓ Correct
(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: (A), (B)

Solution

(A) Work done by external force equals the net potential energy change. Since the ball is pushed slowly (quasi-static), kinetic energy is zero throughout. The work done is:

W_{ext} = (ρ_{w} V g - m g) d = (1000 \times \frac{4}{3} π r^{3} - \frac{22}{7} \times 1 0^{- 3}) \times 10 \times 0.7

= \frac{22}{7} \times 1 0^{- 3} \times \frac{7}{2} \times 7 = 0.077 J

(B) Neglecting viscous force, using energy conservation from depth

d

to surface:

(ρ_{w} V g - m g) d = \frac{1}{2} m v^{2}

\frac{1}{2} \times \frac{22}{7} \times 1 0^{- 3} \times v^{2} = 0.077

v^{2} = 49 \Rightarrow v = 7 m/s

(C) After emerging, the ball rises to height:

H = \frac{v ^{2}}{2 g} = \frac{49}{20} = 2.45 m

This does not match 1.4 m, so (C) is incorrect.

Options (A) and (B) are correct.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · One or more correct options

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JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Fluid Mechanics

Solution

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