Question

As per the figure given below, find the total pressure at the bottom, if the vessel is closed.
Density of liquid 1 = 13.6 $g/cc$
Density of liquid 2 = 5.4 $g/cc$
Density of liquid 3 = 2.5 $g/cc$

(Given $g=9.8m/s^2$)

• A. $97.35\text{kPa}$
• B. $123.68\text{kPa}$
• C. $112.28\text{kPa}$
• D. $145.68\text{kPa}$

Answer 1

The correct option is B $123.68\text{kPa}$

Pressure at the bottom will be the sum of pressure applied by different layers of the liquid with different densities
height of liquid 3 = $h_1=50cm=0.5m$
height of liquid 2 = $h_2=80cm=0.8m$
height of liquid 1 = $h_3=60cm=0.6m$
${\rho }_1=13.6g/cc$ = $13600kg/m^3$
${\rho }_2=5.4g/cc=5400kg/m^3$
${\rho }_3=2.5g/cc=2500kg/m^3$
So applying formula, $P_1+P_2+P_3={\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2+rh{o}_{3}g{h}_3=\text{Total Pressure P}$
Where, $g=\text{acceleration due to gravity}$
$\therefore \text{Total pressure}\left(P\right)=13600\times g\times 0.5+5400\times g\times 0.8+2500\times g\times 0.6$
$P=123676Pa=123.68kPa$