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
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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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