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Physics Question 47 – JEE-MAIN 2025

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A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm2 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 1.5×103 kg/m3. What force one needs to apply on the hinged door so that it does not get opened ? (Acceleration due to gravity = 10 m/s2)

The force required to keep the door closed is equal to the net force exerted by the liquids due to the pressure difference across the door.

● Video Walkthrough
Step 1: Calculate Hydrostatic Pressure on Each Side✦ Active

First, convert the window area to square meters: A=100 cm2=100×(10−2 m)2=10−2 m2. The depth of the window is h=3 m. The density of water is ρw=1000 kg/m3 and the density of the other liquid is ρl=1.5×103 kg/m3=1500 kg/m3. The acceleration due to gravity is g=10 m/s2. Calculate the pressure exerted by each liquid at the depth of the window:

Pw=ρwgh=1000 kg/m3×10 m/s2×3 m=30000 Pa
Pl=ρlgh=1500 kg/m3×10 m/s2×3 m=45000 Pa
Step 2: Determine Net Pressure Difference and Force○ Expand

The net pressure difference across the hinged door is the absolute difference between the pressures on both sides:

ΔP=|Pl−Pw|=|45000 Pa−30000 Pa|=15000 Pa

The force required to keep the door from opening is the product of this net pressure difference and the area of the window:

F=ΔP×A=15000 Pa×10−2 m2=150 N
💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before performing calculations.

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