Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A body of density $1.2 \times 10^{3}\, kg / m ^{3}$ is dropped from rest from a height $1 \,m$ into a liquid of density $2.4 \times 10^{3}\, kg / m ^{3}$. Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is

WBJEEWBJEE 2021

Solution:

$d=1.2 \times 10^{3}, h=1 m, \rho=2.4 \times 10^{3}$
By work Energy Theorem,
$M g(h+d)=B d$
$V\left(1.2 \times 10^{-3}\right) g(1+d)=V\left(2.4 \times 10^{-3}\right) g d $
$\therefore d=1\, m$