A vessel with square cross-section and height of $6\ \text{m}$ is vertically partitioned. A small window of $100\ \text{cm}^2$ with a hinged door is fitted at a depth of $3\ \text{m}$ in the partition wall. One part of the vessel is filled completely with water and the other side is filled with a liquid of density $1.5\times10^3\ \text{kg/m}^3$. What force (in N) needs to be applied on the hinged door so that it does not get opened? (Acceleration due to gravity $= 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 150
At the depth of the window both sides have the same atmospheric pressure at the top, so the net push on the door comes only from the difference in liquid pressure:
$$\Delta P = (\rho_\ell - \rho_w)gh = (1500 - 1000)\times10\times3 = 15000\ \text{Pa}$$
The door is small, so this pressure difference acts over its whole area:
$$F = \Delta P\cdot A = 15000 \times (100\times10^{-4}) = 150\ \text{N}$$
A force of $150\ \text{N}$ must be applied from the water side to keep the door shut.
Solution by Sreeraj P, M.Sc Physics