A table tennis ball has radius $(3 / 2) \times 10^{-2} \mathrm{~m}$ and mass $(22 / 7) \times 10^{-3}…
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.]
The work done in pushing the ball to the depth $d$ is $0.077 \mathrm{~J}$.
If we neglect the viscous force in water, then the speed $v=7 \mathrm{~m} / \mathrm{s}$.
If we neglect the viscous force in water, then the height $H=1.4 \mathrm{~m}$.
The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.
Solution
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} \pi R^3 \times 10 \times 0.7\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 v^2=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, } H=\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 \eta r v \\
& =6 \times \frac{22}{7} \times 10^{-3}\left(\frac{3}{2} \times 10^{-2}\right) \times 7 \\
& =18 \times 11 \times 10^{-5} \mathrm{~N}
\end{aligned}
$
Now,
$
\frac{F_{\text {net }}}{\left(F_v\right)_{\max }}=\frac{500}{9}
$
[4 is correct]