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There is a hole at the side bottom of a big water tank. The area of the hole is $4 {mm}^{2}$ to it a pipe is connected. The upper surface of water is 5m above the hole. The rate of flow of water through the pipe is (in m³/s) $(g = 10 m / s^{2})$

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$4 \times 10^{- 5}$

$28 \times 10^{- 5}$

$4 \times 10^{5}$

$4 \times 10^{- 6}$

answer is A.

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Detailed Solution

Volume flow rate = Area of cross section of the hole= $Q = A \sqrt{2 gh} = 4 \times 10^{- 6} \sqrt{2 \times 10 \times 5} = 4 \times 10^{- 5} m^{3} / s$

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