(D) For the first hole at depth $h/2$ in the liquid of density $\rho$:Using Torricelli's law,$v_1 = \sqrt{2g(h/2)} = \sqrt{gh}$.For the second hole at

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(D) For the first hole at depth $h/2$ in the liquid of density $\rho$:Using Torricelli's law,$v_1 = \sqrt{2g(h/2)} = \sqrt{gh}$.For the second hole at depth $3h/2$ from the surface,it is at a depth $h/2$ within the liquid of density $2\rho$:Applying Bernoulli's theorem between the top surface and the second hole:$P_0 + \rho gh + 2\rho g(h/2) = P_0 + \frac{1}{2}(2\rho)v_2^2$$\rho gh + \rho gh = \rho v_2^2$$2\rho gh = \rho v_2^2$$v_2 = \sqrt{2gh}$.Now,calculating the ratio:$\frac{v_1}{v_2} = \frac{\sqrt{gh}}{\sqrt{2gh}} = \frac{1}{\sqrt{2}}$.

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

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Equal volumes of two immiscible liquids of densities $\rho$ and $2\rho$ are filled in a vessel as shown in the figure. Two small holes are punched at depths $h/2$ and $3h/2$ from the surface of the lighter liquid. If $v_1$ and $v_2$ are the velocities of efflux at these two holes,then $v_1/v_2$ is:

A
$\frac{1}{2\sqrt{2}}$
B
$\frac{1}{2}$
C
$\frac{1}{4}$
D
$\frac{1}{\sqrt{2}}$

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Fluid Mechanics and Surface Tension

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Velocity of Efflux and Torricelli's law

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Equal volumes of two immiscible liquids of densities $\rho$ and $2\rho$ are filled in a vessel as shown in the figure. Two small holes are punched at depths $h/2$ and $3h/2$ from the surface of the lighter liquid. If $v_1$ and $v_2$ are the velocities of efflux at these two holes,then $v_1/v_2$ is:

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$A$ tank is filled up to a height $2H$ with a liquid and is placed on a platform of height $H$ from the ground. The distance $x$ from the ground where a small hole is punched to get the maximum range $R$ is:

$A$ container is filled with water up to a height $h$. $A$ small hole is made at the bottom. What is the ratio of the time taken for the water level to fall from $h$ to $\frac{h}{2}$ to the time taken for it to fall from $\frac{h}{2}$ to $0$?

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