$=\mathrm{mgh}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{gh}$
upthrust on $\mathrm{ball}=\mathrm{V} \rho_{\mathrm{w}} \mathrm{g}$
net force on the ball in water $=\left(\rho_{w}-\rho_{b}\right) V g$
it is constant and upwards
let it sink upto depth d then
$\left(\rho_{m}-\rho_{b}\right) V_{\rho}d=V \rho_{b} g h$
$\Rightarrow \mathrm{d}=\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}-\rho_{\mathrm{b}}} \mathrm{h}$
$\Rightarrow \mathrm{d}=\frac{0.4 \times 10^{3}}{1 \times 10^{3}-0.4 \times 10^{3}} \times 9 \mathrm{cm}=\frac{4}{6} \times 9=6 \mathrm{cm}$



(density of water $=1000\; \mathrm{kgm}^{-3}$ )

$(A)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ $(B)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } > 0$ $(D)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } < 0$