Question

A liquid with varying density is contained in a rectangular container of height $10$$cm$. The density of the liquid at depth h $cm$ below the free surface is given by $\rho ={\rho }_0(1+h)$ , where ${\rho }_0=0.8gmc{m}^{-3}$. Its average density is $\frac{8k}{10}gmc{m}^{-3}$ . The value of ${}^{\prime }{k}^{\prime }$ is

Answer 1

At a depth $h$, density is
$\rho ={\rho }_{\circ }(1+h)$



$\therefore$ Mass of the element is
$dm$ = ${\rho }_0(1+h)Adh$$(\mathrm{where}A\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{cross}-\mathrm{section})$

or $dm=(0.8)(1+h)\left(A\right)h$
As area A will always remain constant in this case.
$\therefore$ Mass of the liquid$\mathrm{is}m=0.8A{\int }_0^{10}(1+h)dh$
$=0.8A\left(10+\frac{{10}^2}{2}\right]=48Agm$
Volume, $V$ = $A\times 10=10Ac{m}^3$
$\therefore {\rho }_{av}=\frac{m}{v}=\frac{48A}{10A}=4.8gc{m}^{-3}\Rightarrow 4.8=\frac{8k}{10}gc{m}^{-3}\Rightarrow k=6$