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(D) Let the top surface of the liquid be point $1$ and the hole be point $2$. Applying Bernoulli's theorem between the top surface (point $1$) and the

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$\sqrt {\frac{{\left( {3H - 4h} \right)g}}{2}} $

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(D) Let the top surface of the liquid be point $1$ and the hole be point $2$. Applying Bernoulli's theorem between the top surface (point $1$) and the hole (point $2$):$P_0 + \rho_1 g \left( \frac{H}{2} \right) + \rho_2 g \left( \frac{H}{2} - h \right) = P_0 + \frac{1}{2} \rho_2 v^2$Here,$\rho_1 = d$,$\rho_2 = 2d$,and $P_0$ is the atmospheric pressure.Substituting the values:$d g \left( \frac{H}{2} \right) + 2d g \left( \frac{H}{2} - h \right) = \frac{1}{2} (2d) v^2$$g \left( \frac{H}{2} + H - 2h \right) = v^2$$g \left( \frac{3H}{2} - 2h \right) = v^2$$v^2 = g \left( \frac{3H - 4h}{2} \right)$$v = \sqrt{\frac{(3H - 4h)g}{2}}$

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