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A small hole of area of cross-section 2 mm2^2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s2^2, the rate of flow of water through the open hole would be nearly

A

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

B

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

C

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

D

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

Step-by-Step Solution

The rate of flow Q=a×vQ = a \times v, where v=2ghv = \sqrt{2gh}. Given a=2 mm2=2×10−6 m2a = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2, h=2 mh = 2 \text{ m}, g=10 m/s2g = 10 \text{ m/s}^2. Q=2×10−6×2×10×2=2×10−6×40≈2×10−6×6.32=12.64×10−6 m3/sQ = 2 \times 10^{-6} \times \sqrt{2 \times 10 \times 2} = 2 \times 10^{-6} \times \sqrt{40} \approx 2 \times 10^{-6} \times 6.32 = 12.64 \times 10^{-6} \text{ m}^3/s.

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