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Q.
A can filled with water is revolved in a vertical circle of radius $4\, m$ and the water does not fall down. The time period for a revolution is about
When a body is revolving in circular motion, it is acted upon by a centripetal force directed towards the centre. Water will not fall, if weight is balanced by centripetal force. Therefore,
$m g =\frac{m v^{2}}{r}$
$\Rightarrow v^{2}=r g \Rightarrow v=\sqrt{r g}$ ...(i)
Circumference of a circle is $2 \pi r$.
Time for a revolution $=\frac{2 \pi r}{v}$
On putting the value of $v$ from Eq. (i), we get
$T=\frac{2 \pi r}{\sqrt{g r}}=2 \pi \sqrt{\frac{r}{g}}$
Given, $r=4 m, g=9.8\, ms ^{-2}$
$\therefore T=2 \pi \sqrt{\frac{4}{9.8}}$
$\Rightarrow T=\frac{4 \pi}{\sqrt{9.8}}=4\, s$