(A) The volume flow rate (quantity of water per second) is given by $Q = A v$, where $A$ is the area of the hole and $v$ is the velocity of efflux.Acc

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(A) The volume flow rate (quantity of water per second) is given by $Q = A v$, where $A$ is the area of the hole and $v$ is the velocity of efflux.According to Torricelli's law, the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.For the square hole: Area $A_1 = L^2$ and depth $h_1 = y$. So, $v_1 = \sqrt{2gy}$.The flow rate $Q_1 = A_1 v_1 = L^2 \sqrt{2gy}$.For the circular hole: Area $A_2 = \pi R^2$ and depth $h_2 = 4y$. So, $v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$.The flow rate $Q_2 = A_2 v_2 = \pi R^2 (2\sqrt{2gy})$.Given that $Q_1 = Q_2$, we have:$L^2 \sqrt{2gy} = 2\pi R^2 \sqrt{2gy}$.Canceling $\sqrt{2gy}$ from both sides, we get $L^2 = 2\pi R^2$.Therefore, $R^2 = \frac{L^2}{2\pi}$, which gives $R = \frac{L}{\sqrt{2\pi}}$.

Class 11 Physics Fluid Mechanics and Surface Tension Velocity of Efflux and Torricelli's law

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$A$ large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then the value of $R$ is

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A
$\frac{L}{\sqrt{2 \pi}}$
B
$2 \pi L$
C
$L \sqrt{\frac{2}{\pi}}$
D
$\frac{L}{2 \pi}$

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Fluid Mechanics and Surface Tension

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Velocity of Efflux and Torricelli's law

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$A$ cylindrical tank has a hole of $1 \, cm^2$ in its bottom. If water is allowed to flow into the tank from a tube above it at the rate of $70 \, cm^3/sec$, then the maximum height up to which water can rise in the tank is .......... $cm$.

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The bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$,it flows out in $t$ seconds. If water is filled to a height $4h$,it will flow out in time:

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Water is filled in a container up to a height of $3\,m$. $A$ small hole of area $A_0$ is punched in the wall of the container at a height of $52.5\,cm$ from the bottom. The cross-sectional area of the container is $A$. If $A_0/A = 0.1$,then $v^2$ is......... $m^2/s^2$ (where $v$ is the velocity of water coming out of the hole).

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$A$ cylindrical tank has a hole of $1 \, cm^2$ in its bottom. If water is allowed to flow into the tank from a tube above it at the rate of $70 \, cm^3/sec$, then the maximum height up to which water can rise in the tank is .......... $cm$.

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