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

 

Answers (1)

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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