JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 23)
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 ?
$$\text { (Acceleration due to gravity }=10 \mathrm{~m} / \mathrm{s}^2 \text { ) }$$
Explanation
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$$\begin{aligned} &\text { in equilibrium }\\ &\begin{aligned} & F_{e x t}+F_w=F_{\ell} \\ & \Rightarrow F_{e x t}=F_{\ell}-F_w \\ & =\left(P_0+\rho_{\ell} g h\right) A-\left(P_0+\rho_w g h\right) A \\ & =\left(\rho_{\ell}-\rho_w\right) g h A \\ & =(1500-1000) \times 10 \times 3 \times\left(100 \times 10^{-4}\right) \\ & =150 \mathrm{~m} \end{aligned} \end{aligned}$$
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