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A small hole of area of cross-section 2 ${mm}^{2}$ present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 $m / s^{2}$ , the rate of flow of water through the open hole would be nearly:

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$6.4 \times 10^{- 6} m^{3} / s$

$12.6 \times 10^{- 6} m^{3} / s$

$8.9 \times 10^{- 6} m^{3} / s$

$2.23 \times 10^{- 6} m^{3} / s$

answer is B.

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Detailed Solution

Velocity of efflux, v = $\sqrt{2 gh}$

Volume flow rate = Av = $A \sqrt{2 gh}$

= (2 $\times 10^{- 6}) {(2 \times 10 \times 2)}^{\frac{1}{2}}$ )

= $4 \sqrt{10} \times 10^{- 6} m^{3} / s$

= 12.6 $\times 10^{- 6} m^{3} / s$

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