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Q.

The area of cross section of a large tank is $0.5 m^{2} .$ It has a narrow opening near the bottom having area of cross section $1 c m^{2}$ . A load of 25kg is applied on the water at the top in the tank. Neglecting the speed of the water, coming out of the opening at the time when the height of water level in the tank is 40cm above the bottom will be _________cm/s
(Take $g = 10 m / s^{2}$ )

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Detailed Solution

By Bernoulli’s theorem,
$P + ρ g h + \frac{1}{2} ρ v^{2} = c o n s t . (P_{0} + \frac{m g}{A}) + ρ g h = P_{0} + \frac{1}{2} ρ v^{2} \frac{25 \times 10}{0.5} + (1000 \times 10 \times 0.4) = \frac{1}{2} \times 1000 \times v^{2} 500 + 4000 = 500 v^{2}$

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