Question 83 of 192

The top of a water tank is open to air and its water level is maintained. It is giving out $0.74{m}^{3}$ water per minute through a circular opening of $2cm$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:

A$2.9m$B$4.8m$C$6.0m$D$9.6m$

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Answer: (B) $4.8m$

Volume of water flowing out$=Av$, where $A$ is area of opening and $v$ is speed of water from the opening.

$\text{Volume}=\pi {r}^{2}v$

$0.74\left(\dfrac{{m}^{3}}{\min }\right)=\left(3.14\times {\left(2\times {10}^{-2}\right)}^{2}\times \sqrt{2\times 10\times H}\right)\dfrac{{m}^{3}}{s}\times \dfrac{60s}{\min }$

$\Rightarrow H={\left(\dfrac{0.74}{60\times 3.14\times 4\times {10}^{-4}}\right)}^{2}\times \dfrac{1}{20}$

$\Rightarrow H=4.8m$
Hence, the height of water in the tank was $4.8m$ from the bottom of the tank.