# A small hole of area of cross sectyion $2 mm ^ 2$ is  present mear the botom  of a fully  filled open tank  of heiht 2 m . Taking $g = 10 m /s ^2$ , the rate of flow of water through the open hole would be nearly :  Option 1) $12.6 \times 10 ^{-6 } m^3 / s$ Option 2) $8.9 \times 10 ^{-6 } m^3 / s$ Option 3) $2.23 \times 10 ^{-6 } m^3 / s$ Option 4) $6.4 \times 10 ^{-6 } m^3 / s$

P Plabita

$A = 2 mm^2 \\\\ P_1 + \rho g h_1 + \frac{1}{2} \rho V_1 ^ 2 = P_2 + \rho g h_2 + \frac{1}{2} \rho V_2 ^ 2 \\\\ V_ 2 = \sqrt {2gh_2 } = \sqrt { 2 \times 10 \times 2 } = \sqrt {40} \\\\ rate \: \: of \: \: flow \: \: = AV \\\\ = (2 \times 10 ^{-6}\times \sqrt {40}) \\\\ 12.64 \times 10 ^ { -6 } m/s$

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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