Physics · Fluid Mechanics

JEE Main 2025 — 2 April, Morning Shift — Question 67

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 $1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ . What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Answer: 150

Numerical answer — enter this value.

Step-by-step solution

$F=\left(P_{2}-P_{1}\right) A$

\begin{aligned} & =\operatorname{gh}\left(\rho_{2}-\rho_{1}\right) A \\ & =10 \times 3 \times 500 \times 100 \times 10^{-4} \\ & =150 \mathrm{~N} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Fluid Mechanics
Topic: Variation of Static Pressure Inside a Liquid