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(D) Let the depths of the two holes from the free surface of the liquid be $y_1$ and $y_2$ respectively. Given that the difference in height is $h$,we

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$h$

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(D) Let the depths of the two holes from the free surface of the liquid be $y_1$ and $y_2$ respectively. Given that the difference in height is $h$,we have $y_2 - y_1 = h$.The velocity of efflux for the top hole is $v_1 = \sqrt{2gy_1}$ and for the bottom hole is $v_2 = \sqrt{2gy_2}$.The force exerted by the liquid jet on the tank is given by $F = \frac{dm}{dt} v = (A \rho v) v = A \rho v^2$,where $A$ is the area of the hole and $\rho$ is the density of the liquid.Since the holes are on opposite sides,the forces act in opposite directions. The net horizontal force $F_{net}$ is $|F_2 - F_1| = |A \rho v_2^2 - A \rho v_1^2|$.Substituting the velocities: $F_{net} = A \rho (2gy_2 - 2gy_1) = 2A \rho g (y_2 - y_1)$.Since $y_2 - y_1 = h$,we get $F_{net} = 2A \rho g h$.Therefore,the net horizontal force is proportional to $h$.

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

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