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The level of water in tank is 5 m high. A hole of area $1 c m^{2}$ is made at the bottom of the tank. The rate of leakage of water from the hole is $(g = 10 m s^{- 2})$

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$10^{- 3} m^{3} s^{- 1}$

$10^{- 4} m^{3} s^{- 1}$

$10 m^{3} s^{- 1}$

$10^{- 2} m^{3} s^{- 1}$

answer is A.

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Detailed Solution

$v = \sqrt{2 g h}$

Since volume flowing per second is

$Q = A v = A \sqrt{2 g h} = (1 \times 10^{- 4}) \times \sqrt{2 \times 10 \times 5}$

$\Rightarrow$ $Q = 10^{- 3} m^{3} s^{- 1}$

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