Question

In a cylindrical container, open to the atmosphere from the top, a liquid is filled upto 10 m depth. Density of the liquid varies with depth from the surface as $\rho \left(h\right)=100+6h^2$ where $h$ is in meter and $\rho$ is in $kg/m^3$. The pressure at the bottom of the container will be: (atmospheric pressure = ${10}^{5}Pa$, $g=10m/se{c}^2$)

• A. $1.7\times {10}^{5}Pa$
• B. $1.4\times {10}^{5}Pa$
• C. $1.6\times {10}^{5}Pa$
• D. $1.3\times {10}^{5}Pa$

Answer 1

The correct option is D $1.3\times {10}^{5}Pa$

$dp=\rho gdh$
${\int }_{P=P_0}^{P=P_f}={\int }_{h=0}^{h=10}(100+6h^2)gdh$
$P_f-P_0={(1000h+20h^3)}_{h=0}^{h=10}$
$\Rightarrow P_f={10}^5+0.3\times {10}^5=1.3\times {10}^5{P}_a$