Question

A stream line body with relative density ${\rho }_1$ falls into air from a height h on the surface of a liquid of relative density ${\rho }_2$, where ${\rho }_2>{\rho }_1$. The time of immersion of the body into the liquid will be:

• A. $t=\sqrt{\frac{2h}{g}}\times \frac{{\rho }_1}{{\rho }_2}$
• B. $t=\sqrt{\frac{2h}{g}}\times \frac{{\rho }_2}{{\rho }_1}$
• C. $t=\sqrt{\frac{2h}{g}}$
• D. $t=\sqrt{\frac{2h}{g}}\times \frac{{\rho }_1}{{\rho }_2-{\rho }_1}$

Answer 1

The correct option is D

$t=\sqrt{\frac{2h}{g}}\times \frac{{\rho }_1}{{\rho }_2-{\rho }_1}$

When the body falls from height h, the velocity by which it hit the surface of liquid $=\sqrt{2gh}$
When the body immersed in liquid of density ${\rho }_2$
Up thrust = $V{\rho }_{2}g$
Weight of body $=V{\rho }_{1}g$
Net retardation $=\frac{V{\rho }_{2}g-V{\rho }_{1}g}{V{\rho }_1}=\frac{({\rho }_2-{\rho }_1)}{{\rho }_1}g$
Time of immersion for body to acquire final velocity zero
$v=v_0+at\Rightarrow 0=\sqrt{2gh}-\left[\frac{{\rho }_2-{\rho }_1}{{\rho }_1}g\right]tt=\frac{\sqrt{2gh}}{g}\frac{{\rho }_1}{{\rho }_2-{\rho }_1}\Rightarrow \sqrt{\frac{2h}{g}}\frac{{\rho }_1}{{\rho }_2-{\rho }_1}$