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.]

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]