Question 219: Properties of Matter

Question

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of $100 \mathrm{~cm}^2$ with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density $1.5 	imes 10^3 \mathrm{~kg} / \mathrm{m}^3$. What force one needs to apply on the hinged door so that it does not get opened ?
$$	ext { (Acceleration due to gravity }=10 \mathrm{~m} / \mathrm{s}^2 	ext { ) }$$

Accepted Answer

Step 1: Find the Pressure on Each Side of Window
At the window's depth, the pressure on each side comes from the liquid above. On one side is water, on the other is a denser liquid.
Step 2: Pressures Acting on Both Sides
The total pressure at depth $h = 3~\mathrm{m}$ in each liquid is:
Pressure from water side: $P_w = P_0 + ho_w g h$
Pressure from dense liquid side: $P_\ell = P_0 + ho_\ell g h$
Where:
$P_0$ = atmospheric pressure (cancels out later)
$ho_w = 1000~\mathrm{kg}/\mathrm{m}^3$
$ho_\ell = 1500~\mathrm{kg}/\mathrm{m}^3$
$g = 10~\mathrm{m}/\mathrm{s}^2$
$h = 3~\mathrm{m}$
Step 3: Net Force on Window
The net force on the window is the difference in pressure between the two liquids, times the window's area:
$F_{ext} = (P_\ell - P_w) \cdot A$
Step 4: Simplifying the Force Expression
The atmospheric pressure, $P_0$, is on both sides, so it cancels out:
$F_{ext} = (ho_\ell g h - ho_w g h) \cdot A$
$= (ho_\ell - ho_w) g h A$
Step 5: Substitute Values
$(ho_\ell - ho_w) = 1500 - 1000 = 500~\mathrm{kg}/\mathrm{m}^3$
$g = 10~\mathrm{m}/\mathrm{s}^2$
$h = 3~\mathrm{m}$
Window area, $A = 100~\mathrm{cm}^2 = 100 	imes 10^{-4}~\mathrm{m}^2 = 0.01~\mathrm{m}^2$
So,

$F_{ext} = 500 	imes 10 	imes 3 	imes 0.01 = 150~\mathrm{N}$
Answer:
The required force to keep the window closed is $150~\mathrm{N}$.

Properties of Matter

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