Question

A body of volume $V$ and density $\rho$ is initially submerged in a liquid of density ${\rho }^{\prime }$. If it lifted through a height $h$ in the liquid, its potential energy will:

• A. increase by $hV\left(\rho -{\rho }^{\prime }\right)g$
• B. decrease by $hV\left(\rho -{\rho }^{\prime }\right)g$
• C. increase by $h{V}_{\rho g/{\rho }^{\prime }}$
• D. decrease by $h{V}_{\rho g/{\rho }^{\prime }}$

Answer 1

The correct option is A increase by $hV\left(\rho -{\rho }^{\prime }\right)g$

Buoyant force$\left(F_B\right)={\rho }^{\prime }vg$

$F_{net}=\rho v_g-F_B$

$=\rho vg-{\rho }^{\prime }g$

$=(\rho -{\rho }^{\prime })vg$

work done in lifting the body upto height 'h'

$\omega =\overrightarrow{F}\cdot \overrightarrow{s}=FS\cos \theta =F_{net}h\cos \left({180}^o\right)$

$\omega =-F_{net}h$

$\omega =-(\rho -{\rho }^{\prime })vg$

since, change in potential energy $\left(\Delta U\right)$:

$\Delta U=-\left(\Delta \omega \right)$

So,

$\Delta U=-\omega =-[-(\rho -{\rho }^{\prime }\left)vg\right]$

$\Rightarrow \Delta U=(\rho -{\rho }^{\prime })vg$

So,

Increase in potential energy by $(\rho -{\rho }^{\prime })vg$.