Equal volumes of two immiscible liquids of densities p and 2p are filled in a vessel as shown in the figure. Two small holes are punched at depth $\frac{h}{2}$ and $\frac{3 h}{2}$ from the surface of lighter liquid. If $v_{1}$ and $v_{2}$ are the velocities of efflux at these holes, then $\frac{v_{1}}{v_{2}}$ will be:

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Solution

We have,

$v_{1} = \sqrt{2 g \times (\frac{h}{2})}$

$v_{1} = \sqrt{g h}$

Now, by using Bernoulli’s theorem

$ρ g h + 2 ρ g (\frac{h}{2}) = \frac{1}{2} (2 ρ) v_{2}^{2}$

$v_{2} = \sqrt{2 g h}$

Now, the value of $\frac{v_{1}}{v_{2}}$

$\frac{v_{1}}{v_{2}} = \frac{\sqrt{g h}}{\sqrt{2 g h}}$

$\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{2}}$

Hence, this is the required solution

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Equal volume of two immiscible liquids of densities $ρ$ and $2 ρ$ are filled in a vessel as shown in the figure. Two small holes are punched at depths $h / 2$ and $3 h / 2$ 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