There is a tank of cross-section area $A_{1}$ with inclined orifice at its bottom with cross-section area $A_{2}$ as shown in the figure. If height of water column in tank is 0.3 m and angle of inclination with vertical $θ = 30^{0}$ and $\frac{A_{1}}{A_{2}} = 2$ , then at this instant find the position where tub is to be placed to collect the water coming out from orifice. $(g = 10 m / s^{2})$

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$\frac{\sqrt{3}}{5} m$

$\frac{2 \sqrt{3}}{5} m$

$\frac{2}{5} m$

$\frac{3 \sqrt{3}}{10} m$

answer is C.

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Detailed Solution

Applying bernouli’s theorem between point on surface of water and point at orifice taking ground as reference,

$P_{a t m} + \frac{1}{2} ρ V_{1}^{2} + ρ g H = P_{a t m} + \frac{1}{2} ρ V_{2}^{2} \Rightarrow V_{2}^{2} - V_{1}^{2} = 2 g H$

$\Rightarrow V_{2}^{2} - (\frac{A_{2}}{A_{1}}) V_{2}^{2} = 2 g H [∵ A_{1} V_{1} = A_{2} V_{2}]$

$\Rightarrow V_{2}^{2} = \frac{2 g H}{1 - {(\frac{A_{2}}{A_{1}})}^{2}}$

Substituting $\frac{A_{2}}{A_{1}} = \frac{1}{2}, H = 0.3 m \Rightarrow V_{2} = 2 \sqrt{2}$

If $θ = 30^{0}$ Range $= \frac{V_{2}^{2} \sin 2 (90 - θ)}{g} = \frac{8 \times \frac{\sqrt{3}}{2}}{10} = \frac{2 \sqrt{3}}{5}$

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