$\therefore$ $\mathrm{mg}=\Delta \mathrm{P} \times \mathrm{A}$
$\Delta P=\frac{m g}{A}=\frac{3 \times 10^{4} \times 10}{120}=2.5 \times 10^{3} P a=2.5 \mathrm{kPa}$
[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]
$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.
$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.
$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.
$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.

