JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2025
- Easy
- Numerical
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 \(=10 \mathrm{~m} / \mathrm{s}^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 150
Step-by-step Solution
Detailed explanation
in equilibrium
\(\begin{aligned} & \mathrm{F}_{\text {ext }}+\mathrm{F}_{\mathrm{w}}=\mathrm{F}_{\ell} \\ & \Rightarrow \mathrm{F}_{\text {ext }}=\mathrm{F}_{\ell}-\mathrm{F}_{\mathrm{w}} \\ & =\left(\mathrm{P}_0+\rho_{\ell} \mathrm{gh}\right) \mathrm{A}-\left(\mathrm{P}_0+\rho_{\mathrm{w}} \mathrm{gh}\right) \mathrm{A} \\ & =\left(\rho_{\ell}-\rho_{\mathrm{w}}\right) \mathrm{ghA} \\ & =(1500-1000) \times 10 \times 3 \times\left(100 \times 10^{-4}\right) \\ & =150 \mathrm{~m}\end{aligned}\)
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