Correct options are (A), (B) and (D)
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
\(\Rightarrow W=\frac{4}{3} \pi R^3 \times 10 \times 0.7\left[1000-\frac{3}{4} \times \frac{10^{-3}}{R^3}\right]\)
\(W=0.077 \mathrm{~J}\)
[A is correct]
\(\Rightarrow\) When ball is released at bottom same work (i.e. \(0.077 \mathrm{~J}\)) is done on ball.
\(\therefore \frac{1}{2} m v^2=0.077\)
\(v=\sqrt{\frac{0.077 \times 2}{\frac{22}{7} \times 10^{-3}}} \)
\(=7 \mathrm{~m} / \mathrm{s}\)
[B is correct]
\(\Rightarrow\) also, \(H=\frac{v^2}{2 g}=\frac{7 \times 7}{2 \times 10}=2.45 \mathrm{~m}\)
[C is incorrect]
\(\Rightarrow\) Net force \(F_{\text {net }}=v \sigma g-v \sigma g=0.11 \mathrm{~N}\)
Also, viscous force is maximum when \(v=7 \mathrm{~m} / \mathrm{s}\)
\(\therefore\left(F_v\right)_{\max } =6 \pi \eta r v\)
\(=6 \times \frac{22}{7} \times 10^{-3}\left(\frac{3}{2} \times 10^{-2}\right) \times 7 \)
\(=18 \times 11 \times 10^{-5} \mathrm{~N}\)
Now,
\(\frac{F_{\text {net }}}{\left(F_v\right)_{\max }}=\frac{500}{9}\)
[D is correct]
