question_answer

     The top of a water tank is open to air and its water level maintained. It is giving out \[0.74\text{ }{{m}^{3}}\] water per minute through a circular opening of 2 cm radius is its wall. The depth of the centre of the opening from the level of water in the tank is close to: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) 6.0 m

B)                                       9.6 m

C) 2.9 m             

D)               4.8 m

Correct Answer: D

Solution :

  \[r\,\,=\,\,2\,cm\] \[a=\pi {{\left( 2\times {{10}^{-2}} \right)}^{2}}=\pi \,\,\times 4\times {{10}^{-}}^{4}{{m}^{2}}\] \[\because \,\,A>>a\] \[\Rightarrow \,\,\,V\,\,=\,\,\sqrt{2gh}\] Volume flow rate = \[Av=0.74{{m}^{3}}/min\] \[=\,\,\frac{0.74}{60}\,{{m}^{3}}/\sec \] \[4\pi \,\,\times \,\,{{10}^{-4}}\,v=\,\,\frac{0.74}{60}\,\] \[v\,\,=\,\,\frac{0.74}{4\pi \,\times \,{{10}^{-4}}\,\times \,\,60}\,\,=\,\frac{0.74\,\times {{10}^{4}}}{4\times 3.14\,\times 60}\] \[v\,\,=\,\,\frac{7400}{12.56\,\times \,60}\,\,=\,\,\sqrt{2gh}\] \[\,\,\frac{740}{75.36}\,\,=\,\,\sqrt{2gh}\] \[h\,\,\approx \,\,4.8\,m\]