Consider a water jar of radius R that has wat | vedclass

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(D) Let $v_1$ be the velocity of water at the hole and $v_2$ be the velocity of water when it hits the ground.Using Torricelli's Law,the velocity of w

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$x = r{\left( {\frac{H}{{H + h}}} \right)^{\frac{1}{4}}}$

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(D) Let $v_1$ be the velocity of water at the hole and $v_2$ be the velocity of water when it hits the ground.Using Torricelli's Law,the velocity of water at the hole is $v_1 = \sqrt{2gH}$.Using the equation of motion for a freely falling body,the velocity at the ground is $v_2 = \sqrt{v_1^2 + 2gh} = \sqrt{2gH + 2gh} = \sqrt{2g(H+h)}$.According to the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A_1 = \pi r^2$ is the area of the hole and $A_2 = \pi x^2$ is the area of the stream at the ground.So,$\pi r^2 \sqrt{2gH} = \pi x^2 \sqrt{2g(H+h)}$.Dividing both sides by $\pi \sqrt{2g}$,we get $r^2 \sqrt{H} = x^2 \sqrt{H+h}$.Squaring both sides,$r^4 H = x^4 (H+h)$.Therefore,$x^4 = r^4 \left( \frac{H}{H+h} \right)$,which gives $x = r \left( \frac{H}{H+h} \right)^{\frac{1}{4}}$.

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