Physics - Fluid Question with Solution | TestHub

$\frac{{\rho }_1}{{\rho }_2}=\frac{1}{2}$ i.e., in equilibrium, block is half immersed in water. Let $h^{\prime }$ be the increase in level when the block is pressed by an amount $h$. Then $\left(A-l^2\right)h^{\prime }=l^2.h$ Substituting the values, we get $h^{\prime }=0.5h$ We have to immerse further $\frac{1}{2}$. Hence $\frac{1}{2}+h+h^{\prime }=1.5h$ or $h=\frac{1}{3}$ When the block is depressed by $h$, extra upthrust $F=\left(h+h^{\prime }\right){\rho }_2{l}^2.g=1.5l^2{\rho }_{2}gh$ Work has to be done against this upthrust. Hence $W={\int }_0^{1/3}Fdh={\int }_0^{1/3}\left(1.5l^2{\rho }_{2}gh\right)dh$ Substituting the values, we get $W=6.75\mathrm{J}$