$\mathrm{mgh}=\frac{1}{2} \mathrm{mu}^{2} \quad$ or $\quad \mathrm{u}=\sqrt{2 \mathrm{gh}}$

Let $\quad$ volume of the body $=V$

mass of the body $=\mathrm{Vd}$

weight of the body $=\mathrm{Vdg}$

Mass of liquid displaced $= VD$

Weight of liquid displaced $=\mathrm{VDg}$

Net upward force $=\mathrm{VDg}-\mathrm{Vdg}$

$=\mathrm{Vg}(\mathrm{D}-\mathrm{d})$

Retardation $=\frac{\text { net weight }}{\text { mass }}$

$=\frac{V(D-d) g}{V d}=\left(\frac{D-d}{d}\right) g$

Acceleration $a=-\left(\frac{D-d}{d}\right) g$

Final velocity, v in the liquid when the body is instantaneously at rest is zero. Let the time taken be $t.$

$\text { Now } \quad \mathrm{v}=\mathrm{u}+a \mathrm{t}$

$0=\sqrt{2 \mathrm{gh}}-\left(\frac{\mathrm{D}-\mathrm{d}}{\mathrm{d}}\right) \mathrm{gt},\left(\frac{\mathrm{D}-\mathrm{d}}{\mathrm{d}}\right) \mathrm{gt}=\sqrt{2 \mathrm{gh}}$

$\therefore \quad t=\left[\frac{d}{D-d}\right] \sqrt{\frac{2 h}{g}}$