A cylindrical tank is filled with water to level of 3 m. A hole is opened at 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. The square of the speed with which water is coming out from the orifice is (Take g = 10 m s^-2)

50 m² s^-²

40 m² s^-2

51.5 m² s^-2

50.5 m² s^-2

Solution

50 m² s^-²

Suppose A be the area of cross section of tank, a be the area of hole, v_e be the velocity of
efflux, h be the height of liquid above the hole,

Let v be the speed with which the level decreases in the container. Using equation of
continuity, we get

av_e = Av

⇒ $v = \frac{{av}_{e}}{A}$

Using Bernoulli's theorem, we have

P₀ + hρg + $\frac{1}{2} {ρν}^{2}$ = P₀ + $\frac{1}{2} ρ v_{e}^{2}$

ρ = density of fluid

P = pressure

⇒ hρg + $\frac{1}{2} ρ {(\frac{{av}_{e}}{A})}^{2} = \frac{1}{2} {ρv}_{e}^{2}$

⇒ $v_{e}^{2}$ = $\frac{2 hg}{1 - (\frac{a^{2}}{A^{2}})}$

= $\frac{2 \times (3 - 0.525)}{1 - {(0.1)}^{2}}$

⇒ $v_{e}^{2}$ = 50 m² s^-2

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