Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = l0 m/s2)

A
50 $m^{2} / s^{2}$
B
50.5 $m^{2} / s^{2}$
C
51 $m^{2} / s^{2}$
D
52 $m^{2} / s^{2}$

Solution:

Let A = cross-section of tank
a = cross-section hole
V = velocity with which level decreases
v = velocity of effiux

From equation of continuity av = AV
$\Rightarrow V = \frac{av}{A}$

By using Bernoulli's theorem for energy per unit
volume
Energy per unit volume at point A = Energy per unit volume at point B
$P + ρgh + \frac{1}{2} {ρv}^{2} = P + 0 + \frac{1}{2} {ρv}^{2}$

$v^{2} = \frac{2 gh}{1 - {(\frac{a}{A})}^{2}}$

= $\frac{2 \times 10 \times (3 - 0.525)}{1 - {(0.1)}^{2}} = 50 {(m / \sec)}^{2}$

Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = l0 $m / s^{2}$ )

Solution:

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