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 $\rho (h) = 100 + 6h^2$ where $h$ is in meter and $\rho$ is in $kg/m^3.$ The pressure at the bottom of the container will be : $($ atmosphere pressure $= 10^5\,\, Pa, \,g = 10\, m/sec^2)$
Diffcult
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$p=\int_{x}^{E} \rho g d y+p_{0}$
$=10 \times \int_{0}^{10}\left(100+6 \mathrm{R}^{2}\right) \mathrm{d} \mathrm{h}+10^{5}$
$=10^{4}+2 \times 10^{4}+10^{5}=1.3 \times 10^{5} \mathrm{Pa}$
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