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 09-Jan-2019 Evening]

A)  6.0 m   

B)                   9.6 m

C)  2.9 m             

D)       4.8 m

Correct Answer: D

Solution :

[d]

\[r\,\,=\,\,2\,cm\]

\[a=\pi {{\left( 2\times {{10}^{-2}} \right)}^{2}}=\pi \,\,\times 4\times {{10}^{-}}^{4}{{m}^{2}}\]

\[\because \,\,A>>a\]

\[\Rightarrow \,\,\,V\,\,=\,\,\sqrt{2gh}\]

Volume flow rate = \[Av=0.74{{m}^{3}}/min\]

\[=\,\,\frac{0.74}{60}\,{{m}^{3}}/\sec \]

\[4\pi \,\,\times \,\,{{10}^{-4}}\,v=\,\,\frac{0.74}{60}\,\]

\[v\,\,=\,\,\frac{0.74}{4\pi \,\times \,{{10}^{-4}}\,\times \,\,60}\,\,=\,\frac{0.74\,\times {{10}^{4}}}{4\times 3.14\,\times 60}\]

\[v\,\,=\,\,\frac{7400}{12.56\,\times \,60}\,\,=\,\,\sqrt{2gh}\]

\[\,\,\frac{740}{75.36}\,\,=\,\,\sqrt{2gh}\]

\[h\,\,\approx \,\,4.8\,m\]