Question

A body of volume $V$ and density $\rho$ is raised through height $h$, in a liquid of density $\sigma (\sigma <\rho )$. The increment in potential energy of the body is (Given acceleration due to gravity $=g$):

• A. $V\rho gh$
• B. $V\sigma gh$
• C. $V(\rho +\sigma )gh$
• D. $V(\rho -\sigma )gh$

Answer 1

Answer: (D)

$V(\rho -\sigma )gh$

Here, a body of volume V and density $\rho$ is raised through height h, in a liquid of density $\sigma$. When the body is raised in liquid equal amount of liquid get displaced.
Let, m be the mass of body, $m=\rho V$ and P is the potential energy of body, $P=mgh=\rho Vgh$.
Now, equal mass of liquid displaced is $m^{\prime }=\sigma V$
Hence, P' be loss in potential energy of liquid of density $\sigma$ when it replaces the volume V when liquid of density $\rho$ has raised, $P^{\prime }=\sigma Vgh$.
the increase in potential energy is
$P=P-P^{\prime }=\rho Vgh-\sigma Vgh$
$P=V(\rho -\sigma )gh$