A vessel of area of cross-section A has liqui | vedclass

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(A) According to Torricelli's law,the velocity of efflux at height $h$ is $v = \sqrt{2gh}$.The rate of volume flow out of the hole is $dV/dt = a \cdot

Vedclass · Accepted Answer

$\frac{A}{a}\sqrt {\frac{2}{g}} [\sqrt {{H_1}} - \sqrt {{H_2}} ]$

Vedclass · Answer

(A) According to Torricelli's law,the velocity of efflux at height $h$ is $v = \sqrt{2gh}$.The rate of volume flow out of the hole is $dV/dt = a \cdot v = a \sqrt{2gh}$.Since the volume of liquid in the vessel decreases as the level drops,we have $dV = -A \cdot dh$ (where $dh$ is the change in height).Equating the two expressions for $dV/dt$: $-A \cdot dh = a \sqrt{2gh} \cdot dt$.Rearranging for $dt$: $dt = -\frac{A}{a \sqrt{2g}} \cdot h^{-1/2} \cdot dh$.Integrating from $H_1$ to $H_2$ to find the total time $t$: $t = \int_{0}^{t} dt = -\frac{A}{a \sqrt{2g}} \int_{H_1}^{H_2} h^{-1/2} \cdot dh$.$t = \frac{A}{a \sqrt{2g}} \int_{H_2}^{H_1} h^{-1/2} \cdot dh = \frac{A}{a \sqrt{2g}} [2\sqrt{h}]_{H_2}^{H_1}$.$t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_1} - \sqrt{H_2}]$.

$A$ vessel of area of cross-section $A$ has liquid to a height $H$. There is a hole at the bottom of the vessel having area of cross-section $a$. The time taken to decrease the level from $H_1$ to $H_2$ will be:

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$A$ completely filled water tank of height $h$ has a hole at the bottom. The total pressure at the bottom is $4H$ and the atmospheric pressure is $H$. The velocity of water flowing out of the hole is $(\rho = \text{density of water})$

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