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There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density p. The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will have to be applied on the tank to keep it in equilibrium is:

Option: 1
$\mathrm{ { gh \rho a }}$

Option: 2
$\mathrm{\frac{2 g h}{\rho a}}$

Option: 3
$\mathrm{ { 2 \rho agh }}$

Option: 4
$\mathrm{\frac{\rho g h}{a}}$

Answers (1)

best_answer

Rate of change of momentum $\mathrm{=\frac{\mathrm{dp}}{\mathrm{dt}}}$

$\mathrm{ \begin{aligned} & \Rightarrow \mathrm{V} \frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{V} \rho \mathrm{aV}=\mathrm{aV}^2 \rho \\\\ & \text { and } \mathrm{V}=\sqrt{2 \mathrm{gh}^1} \\\\ & \Delta \mathrm{F}=\rho \mathrm{a}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right) \end{aligned} }$

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Sanket Gandhi

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