(A) The rate of flow of water (volume flow rate) is given by $Q = A \times v$,where $A$ is the area of the hole and $v$ is the velocity of efflux.Acco

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(A) The rate of flow of water (volume flow rate) is given by $Q = A \times v$,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 $d$ is $v = \sqrt{2gd}$.For the square hole at depth $h$: Area $A_1 = l^2$,velocity $v_1 = \sqrt{2gh}$. Thus,$Q_1 = l^2 \sqrt{2gh}$.For the circular hole at depth $4h$: Area $A_2 = \pi r^2$,velocity $v_2 = \sqrt{2g(4h)} = 2\sqrt{2gh}$. Thus,$Q_2 = \pi r^2 (2\sqrt{2gh})$.Given that the rates of flow are equal,$Q_1 = Q_2$:$l^2 \sqrt{2gh} = 2\pi r^2 \sqrt{2gh}$$l^2 = 2\pi r^2$$r^2 = \frac{l^2}{2\pi}$$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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$A$ square hole of side length $l$ is made at a depth of $h$ and a circular hole of radius $r$ is made at a depth of $4h$ from the surface of water in a water tank kept on a horizontal surface. If $l << h$ and $r << h$,and the rate of water flow from the holes is the same,then $r$ is equal to:

A
$\frac{l}{\sqrt{2\pi}}$
B
$\frac{l}{\sqrt{3\pi}}$
C
$\frac{l}{3\pi}$
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

$A$ cylindrical container containing water has a small hole at a height of $H = 8 \text{ cm}$ from the bottom and at a depth of $h = 2 \text{ cm}$ from the top surface of the liquid. The horizontal distance (range) travelled by the water before it hits the ground is:

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The top of a water tank is open to air and its water level is maintained. It is discharging $0.74\,m^3$ of water per minute through a circular opening of $2\,cm$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to........ $m$

MediumJEE MAIN 2019

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$A$ cylindrical tank has a hole of area $2 \,cm^2$ at its bottom. If water is poured into the tank from a tube above it at the rate of $100 \,cm^3 \,s^{-1}$, then the maximum height up to which water can rise in the tank is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $cm$)

MediumAP EAMCET 2019

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$A$ water tank kept on the ground has an orifice of $2 \,mm$ diameter on the vertical side. What is the minimum height of the water above the orifice for which the output flow of water is found to be turbulent (in $\,cm$)? (Assume, $g=10 \,m/s^2, \rho_{\text{water}}=10^3 \,kg/m^3$, viscosity $=1$ centi-poise)

MediumTS EAMCET 2019

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$A$ tank is filled up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then:

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$A$ square hole of side length $l$ is made at a depth of $h$ and a circular hole of radius $r$ is made at a depth of $4h$ from the surface of water in a water tank kept on a horizontal surface. If $l << h$ and $r << h$,and the rate of water flow from the holes is the same,then $r$ is equal to:

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$A$ cylindrical container containing water has a small hole at a height of $H = 8 \text{ cm}$ from the bottom and at a depth of $h = 2 \text{ cm}$ from the top surface of the liquid. The horizontal distance (range) travelled by the water before it hits the ground is:

The top of a water tank is open to air and its water level is maintained. It is discharging $0.74\,m^3$ of water per minute through a circular opening of $2\,cm$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to........ $m$

$A$ cylindrical tank has a hole of area $2 \,cm^2$ at its bottom. If water is poured into the tank from a tube above it at the rate of $100 \,cm^3 \,s^{-1}$, then the maximum height up to which water can rise in the tank is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $cm$)

$A$ tank is filled up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then:

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