(A) The velocity of efflux $(v)$ through a hole at a depth $H$ below the free surface is given by Torricelli's law: $v = \sqrt{2gH}$.The volume flow r

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(A) The velocity of efflux $(v)$ through a hole at a depth $H$ below the free surface is given by Torricelli's law: $v = \sqrt{2gH}$.The volume flow rate $(Q)$ is the product of the area of the hole $(a)$ and the velocity of efflux $(v)$:$Q = a \times v = a \sqrt{2gH}$Given:$Q = 2 \times 10^{-3} \,m^3/s$$a = 5 \,cm^2 = 5 \times 10^{-4} \,m^2$$g = 10 \,ms^{-2}$Substituting the values into the equation:$2 \times 10^{-3} = 5 \times 10^{-4} \times \sqrt{2 \times 10 \times H}$Divide both sides by $5 \times 10^{-4}$:$4 = \sqrt{20H}$Squaring both sides:$16 = 20H$$H = \frac{16}{20} \,m = 0.8 \,m = 80 \,cm$.

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

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$A$ cylindrical tank having a large diameter is filled with water to a height $H$. $A$ hole of cross-sectional area $5 \,cm^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3} \,m^3 \,s^{-1}$, then the value of $H$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,cm$)

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Fluid Mechanics and Surface Tension

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Velocity of Efflux and Torricelli's law

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There are two identical small holes on the opposite sides of a tank filled with a liquid. The tank is open at the top. The difference in height between the two holes is $h$. As the liquid comes out of the two holes,the tank will experience a net horizontal force proportional to

$A$ 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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