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A can filled with water is revolved in a vertical circle of radius $r$ with constant speed and water just does not fall down. The time period of revolution is $(g=$ acceleration due to gravity)
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Verified Answer
The correct answer is:
$2 \pi \sqrt{\frac{r}{g}}$
For water not to fall off the can, then centrifugal force must balance the weight of water,
$\therefore \frac{m v^2}{r} \geq m g$
Minimum speed of the can so that water does not fall down is given by,
$v=\sqrt{r g}$
Time period of revolution,
$T=\frac{2 \pi r}{v}=2 \pi \sqrt{\frac{r}{g}}$
$\therefore \frac{m v^2}{r} \geq m g$
Minimum speed of the can so that water does not fall down is given by,
$v=\sqrt{r g}$
Time period of revolution,
$T=\frac{2 \pi r}{v}=2 \pi \sqrt{\frac{r}{g}}$
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