JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 18)
The top of a water tank is open to air and its water level is mainted. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the center of the opening from the level of water in the tank is close to :
6.0 m
4.8 m
9.6 m
2.9 m
Explanation
Here water level is kept same all the time. So, the amount of water remove from the hole put in the tank the top to keep the water level same.
$$ \therefore $$ Inflow volume role = outflow volume
$$ \Rightarrow $$ $${{0.74} \over {60}} = Av$$
$$ \Rightarrow $$ $${{0.74} \over {60}} = \pi {r^2}\left( {\sqrt {2gh} } \right)$$
$$ \Rightarrow $$ $${{0.74} \over {60}} = \pi \left( {4 \times {{10}^{ - 4}}} \right) \times \sqrt {2gh} $$
$$ \Rightarrow $$ $$\sqrt {2gh} $$ = $${{740} \over {24\pi }}$$
$$ \Rightarrow $$ 2gh = $${{740 \times 740} \over {24 \times 24 \times {\pi ^2}}}$$
$$ \Rightarrow $$ h = 4.8 m
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