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Physics Question on mechanical properties of fluid

A small hole of area of cross-section 2 mm22 \,mm^2 is present near the bottom of a fully filled open tank of height 2 m2 \,m. Taking g=10 m/s−2g = 10 \,m/s^{-2}. the rate of flow of water through the open hole would be nearly:

A

2.23×10−6m3/s−12.23 \times 10^{-6}m^3/s^{-1}

B

6.4×10−6m3/s−16.4 \times 10^{-6}m^3/s^{-1}

C

12.6×10−6m3/s−112.6 \times 10^{-6}m^3/s^{-1}

D

8.9×10−6m3/s−18.9 \times 10^{-6}m^3/s^{-1}

Answer

12.6×10−6m3/s−112.6 \times 10^{-6}m^3/s^{-1}

Explanation

Solution

Rate of flow liquid
Q=au=a2ghQ =au = a\sqrt{2gh}
=2×10−6m2×2×10×2m/s=2\times10^{-6} m^{2} \times\sqrt{2\times10\times2} m/s
=2×2×3.14×10−6m3/s−1= 2\times2 \times3.14 \times10^{-6} m^{3}/s ^{-1}
=12.56×10−6m3/s−1= 12.56\times10^{-6} m^{3}/s ^{-1}
=12.6×10−6m3/s−1= 12.6 \times10^{-6} m^{3} /s ^{-1}