Two cylindrical vessels A and B of different areas of cross-section kept on same horizontal plane are filled…
- $1: 1$
- $1: 3$
- $1: 9$
- $1: 6$
Solution

$\begin{aligned} & \mathrm{V}_1=3 \mathrm{~V}_2 \\ \Rightarrow & \mathrm{~A}_1 \mathrm{~h}=3 \mathrm{~A}_2 \mathrm{~h} \\ \therefore & \mathrm{~A}_1=3 \mathrm{~A}_2 \end{aligned}$ $\therefore \quad$ Pressure at the bottom of the vessel is $\begin{aligned} & \mathrm{P}=\mathrm{P}_0+\delta \mathrm{gh}=\text { constant } \\ & \therefore \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{1}{1}=1: 1 \end{aligned}$
Asked in: AP EAMCET 2024 (22 May Shift 2)
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