$\rho v_{2}-\rho_{\omega}\left(v_{1}+v_{2}\right)=7.5$ (in water)
$\rho_{\omega}\left(v_{1}+v_{2}\right)=2.5$
$\frac{\rho_{\omega}\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}{\rho\left(\mathrm{v}_{2}\right)}=\frac{2.5}{10}=\frac{1}{4}$
$\frac{1000}{8000}\left(\frac{v_{1}}{v_{2}}+1\right)=\frac{1}{4}$
$\frac{v_{1}}{v_{2}}=1$
[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 $500 / 9$.

(ignore viscosity of air)
(given atmospheric pressure $P_{A}=1.01 \times 10^{5}\,Pa$, density of water $\rho_{ w }=1000\,kg / m ^{3}$ and gravitational acceleration $g=10\,m / s ^{2}$ )