Two identical tall jars are filled with water to the brim. The first jar has a small hole on the side wall at a depth $h / 3$ and the second jar has a small hole on the side wall at a depth of $2 h / 3$ , where h is the height of the jar. The water issuing out from the first jar falls at a distance $R_{1}$ from the base and the water issuing out from the second jar falls at a distance $R_{2}$ from the base. The correct relation between $R_{1} a n d R_{2}$ is

R_{1} > R_{2}

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R_{1} < R_{2}

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R_{2} = 2 \times R_{1}

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R_{1} = R_{2}

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Solution

The correct option is D $R_{1} = R_{2}$
At any height h from surface, velocity $V = \sqrt{2 g h}$
Time of flight $t = \sqrt{\frac{2 (H - h)}{g}}$
where $H$ is the total height of jar.
$∴$ Range $R = V \times t$
$R = 2 \sqrt{h (H - h)}$
Given, $h_{1} = \frac{h}{3} a n d H = h$
$\Rightarrow$ $R_{1} = 2 \sqrt{\frac{h}{3} (h - \frac{h}{3})}$ $= \frac{2 \sqrt{2} h}{3}$
For $h_{2} = \frac{2 h}{3} a n d H = h$
$\Rightarrow$ $R_{2} = 2 \sqrt{\frac{2 h}{3} (h - \frac{2 h}{3})}$ $= \frac{2 \sqrt{2} h}{3}$
$∴ R_{1} = R_{2}$

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