JEE Advanced 2024Paper 2

JEE Advanced 2024 · Paper 2 · Physics · Mechanical Properties of Fluids

A table tennis ball has radius (3 / 2) × 10^ -2 ~m and mass (22 /...

A table tennis ball has radius

(3 / 2) \times 10^{-2} \mathrm{~m}

and mass

(22 / 7) \times 10^{-3} \mathrm{~kg}

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

d=0.7 \mathrm{~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:

\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}

, density of water

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

, viscosity of water

=1 \times 10^{-3} \mathrm{~Pa}

-s.]

Options

AThe work done in pushing the ball to the depth $d$ is $0.077 \mathrm{~J}$ .
BIf we neglect the viscous force in water, then the speed $v=7 \mathrm{~m} / \mathrm{s}$ .
CIf we neglect the viscous force in water, then the height $H=1.4 \mathrm{~m}$ .
DThe ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$ .

A. The work done in pushing the ball to the depth $d$ is $0.077 \mathrm{~J}$ .

Work done in pushing the ball W=(v \rho g) d-(v

\sigma

g) d Where,

\rho \rightarrow

Density of water

\sigma \rightarrow

Density of ball

\begin{aligned}& \Rightarrow W=\frac{4}{3} \piR^3 \times10 \times0.7\left[1000-\frac{3}{4} \times \frac{10^{-3}}{R^3}\right]\\ & W=0.077 \mathrm{~J} \end{aligned}

[

1 is correct] \Rightarrow When ball is released at bottom same work (i.e.

0.077 \mathrm{~J}

) is done on ball.

\begin{aligned}& \therefore \frac{1}{2}m v²=0.077 \\ & v=\sqrt{\frac{0.077 \times 2}{\frac{22}{7} \times 10^{-3}}}\\ & =7 \mathrm{~m}/ \mathrm{s} \end{aligned}

[

2 is correct] \Rightarrow

\text{ also, }

\frac{v^2}{2 g}=\frac{7 \times 7}{2 \times 10}=

2.45

\mathrm{~m} [

3 is incorrect] \Rightarrow Net force

F_{\text{net }}=v \sigma g-v \sigma g=0.11 \mathrm{~N}

Also, viscous force is maximum when

v=7 \mathrm{~m} / \mathrm{s}

\begin{aligned}& \therefore\left(F_v\right)_{\max }=6 \pi \etar v \\ & =6 \times \frac{22}{7} \times10^{-3}\left(\frac{3}{2} \times10^{-2}\right) \times7 \\ & =18 \times11 \times10^{-5} \mathrm{~N} \end{aligned}

$$Now, \frac{F_{\text{net }}}{\left(F_v\right)_{\max }}=\frac{500}{9} [4 is correct]