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A tank, which is open at the top, contains a liquid up to a height \( \boldsymbol{H} . \mathbf{A} \) small hole is made in the side of a tank at a distance y below the liquid surface. The liquid emerging from the hole lands at a distance \( x \) from the tank. his question has multiple correct options

1. If \( y \) is increased from zero to \( H, x \) will first increase and then decrease
2. x is maximum for \( y=H / 2 \).
3. The maximum value of \( x \) is \( H \)
4. The maximum value of \( x \) will depend on the density of the density of the liquid.

Correct answer: x is maximum for \( y=H / 2 \).

Solution

The jet leaves the hole with speed \(\sqrt{2gy}\) and falls a vertical distance \(H-y\), so the time of flight is \(\sqrt{2(H-y)/g}\). Thus \(x = \sqrt{2gy}\,\sqrt{2(H-y)/g} = 2\sqrt{y(H-y)}\), which is maximized when \(y=H/2\).

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