A tank of oil has height of $4 \mathrm{~m}$ and density of $850 \mathrm{~kg} \mathrm{~m}^{-3}$. The gauge…
A tank of oil has height of $4 \mathrm{~m}$ and density of $850 \mathrm{~kg} \mathrm{~m}^{-3}$. The gauge pressure at the bottom of the tank is (1 atm $=10^5 \mathrm{~Pa}$, Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$34 \mathrm{kPa}$
$384 \mathrm{kPa}$
$284 \mathrm{kPa}$
$200 \mathrm{kPa}$
Solution
Tank of oil has height, $h=4 \mathrm{~m}$ Density, $\mathrm{P}=850 \mathrm{~kg} / \mathrm{m}^3$ Gauge pressure, $P=\rho \mathrm{gh}$ $=850 \times 10 \times 4=34 \mathrm{kPa}$