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Properties of Matter

Section: Physics

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Question : 2 of 47

Marks: +1, -0

A table tennis ball has radius

\left(\frac{3}{2}\right) \times 10^{-2} \text{m}

\left(\frac{22}{7}\right) \times 10^{-3} \text{kg}

. It is slowly pushed down into a swimming pool to a depth of

d=0.7 \text{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?

\pi = \frac{22}{7}, g = 10 \text{ms}^{-2}

, density of water

= 1 \times 10^3 \text{kg} \text{m}^{-3}

, viscosity of water

= 1 \times 10^{-3} \text{Pa}-\text{s}

[JEE Adv 2024 P2]

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 $v=7 \text{m/s}$ .
If we neglect the viscous force in water, then the height $H=1.4 \text{m}$ .
The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

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