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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2025 · 2 April · Morning Shift · Physics · Mechanical Properties of Fluids

A vessel with square cross-section and height of 6 m is vertically...

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm2100 \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 1.5×103 kg/m31.5 \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 m/s2=10 \mathrm{~m} / \mathrm{s}^2)

Correct answer

150

Step-by-step solution

in equilibriumFext +Fw=Fℓ⇒Fext =Fℓ−Fw=(P0+ρℓgh)A−(P0+ρwgh)A=(ρℓ−ρw)ghA=(1500−1000)×10×3×(100×10−4)=150m\begin{aligned} & \mathrm{F}_{\text{ext }}+\mathrm{F}_{\mathrm{w}}=\mathrm{F}_{\ell} \\ & \Rightarrow \mathrm{F}_{\text{ext }}=\mathrm{F}_{\ell}-\mathrm{F}_{\mathrm{w}} \\ & =\left(\mathrm{P}_0+\rho_{\ell} \mathrm{gh}\right) \mathrm{A}-\left(\mathrm{P}_0+\rho_{\mathrm{w}} \mathrm{gh}\right) \mathrm{A} \\ & =\left(\rho_{\ell}-\rho_{\mathrm{w}}\right) \mathrm{ghA} \\ & =(1500-1000) \times 10 \times 3 \times\left(100 \times 10^{-4}\right) \\ & =150 \mathrm{ m}\end{aligned}
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