A hemispherical tank of diameter 4m is filled with water. A very small hole is punched at 1m above bottom as shown. Center of hemispherical tank is considered as origin, find x – co – ordinate in meter, when water strike the horizontal surface. $(g = 10 m / s^{2})$ (round off to nearest integer)

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answer is 3.

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Detailed Solution

$\begin{array}{l} p_{0} + ρg (1) + 0 = p_{0} + 0 + \frac{1}{2} {ρv}^{2} \\ \sqrt{2 g} = v \\ 1 = vcos θt + \frac{1}{2} {gt}^{2} \\ 1 = \sqrt{2 g} (\frac{1}{2}) t + 5 t^{2} \\ 1 = \sqrt{5} t + 5 t^{2} \\ 5 t^{2} + \sqrt{5} t - 1 = 0 \\ t = \frac{- \sqrt{5} \pm \sqrt{5 + 20}}{10} \\ t = \frac{- \sqrt{5} \pm 5}{10} = \frac{5 - \sqrt{5}}{10} \\ R = vsin θt \\ = \sqrt{2 g} (\frac{\sqrt{3}}{2}) \cdot (\frac{5 - \sqrt{5}}{10}) \\ = \sqrt{5} \sqrt{3} (\frac{5 - \sqrt{5}}{10}) \\ = \sqrt{3} (\frac{\sqrt{5} - 1}{2}) \\ = \frac{\sqrt{15}}{2} - \frac{\sqrt{3}}{2} \\ x = R + \sqrt{3} \end{array}$
$\begin{array}{l} = \frac{\sqrt{15}}{2} - \frac{\sqrt{3}}{2} + \sqrt{3} \\ = \frac{\sqrt{15}}{2} + \frac{\sqrt{3}}{2} \\ = \frac{\sqrt{3} (\sqrt{5} + 1)}{2} \\ = \frac{1.7}{2} (2.2 + 1) \\ = \frac{1.7}{2} \times 3.2 \\ = 1.7 \times 1.6 \\ = 2.72 = 3 \end{array}$