A cylindrical, tank is filled with water to a level of 3m. A hole is opened at a height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1. Find the square of the velocity with which water is coming out. (g = 10 m/sec²)

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$50.5 m^{2} / \sec^{2}$

$40 m^{2} / \sec^{2}$

$51.5 m^{2} / \sec^{2}$

$50 m^{2} / \sec^{2}$

answer is A.

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Detailed Solution

Let velocity with which level decreases is v₁ and velocity of efflux is v₂then
From equation of continuity ${Av}_{1} = {av}_{2} \Rightarrow v_{1} = \frac{a}{A} v_{2}$
From Bernoulli's theorem
$P + \frac{1}{2} {ρv}_{1}^{2} + ρgh = P + 0 + \frac{1}{2} {ρv}_{2}^{2} v_{2}^{2} = \frac{2 gh}{1 - {(\frac{v_{1}}{v_{2}})}^{2}} = \frac{2 gh}{1 - {(\frac{a}{A})}^{2}} = \frac{2 \times 10 \times (3 - 0.525)}{1 - {(0.1)}^{2}} = 50 m^{2} / \sec^{2}$