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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 $1 {cm}^{2}$ . A load of $25 kg$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of water, coming out of the opening at the time when water level in the tank is $40 cm$ above the bottom, will be ${cms}^{- 1} [Take g = 10 {ms}^{- 2}]$

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answer is 300.

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Detailed Solution

Using Bernoulli's principle :

$P_{A} + \frac{1}{2} {ρV}_{A}^{2} = P_{B} + \frac{1}{2} ρ {V_{B}}^{2}$

$P_{0} + \frac{250}{0.5} + (0.4 \times 1000 \times 10) + \frac{1}{2} {ρ (0)}^{2} = P_{0} + \frac{1}{2} ρ {V_{B}}^{2}$

$4500 = \frac{1}{2} (1000) {V_{B}}^{2}$

$V_{B} = 3 m / s = 300 cm / s$

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