A can filled with water is revolved in a vertical circle of radius 4m and the water just does not fall down. The time period of revolution will be

The correct answer/option is 4 sec.

Let's understand the physics of a can filled with water revolving in a vertical circle. For the water not to fall down when the can is at the highest point of the vertical path, the centrifugal force acting outwards (in the co-rotating frame of reference) must be at least equal to or greater than the weight of the water acting downwards. Alternatively, in the ground frame, the centripetal acceleration at the top must be at least equal to the acceleration due to gravity $g$ so that the normal force between the bottom of the can and the water remains greater than or equal to zero.

Thus, the limiting condition for the water to just not fall out at the top of the vertical circle of radius $R$ is given by:
$\frac{v^2}{R}=g$

Using the relationship between linear velocity $v$ and angular velocity $\omega$, where $v=\omega R$, we can write:
${\omega }^{2}R=g$

Since the angular velocity is related to the time period of revolution $T$ by $\omega =\frac{2\pi }{T}$, we substitute this into the equation:
${\left(\frac{2\pi }{T}\right)}^{2}R=g$

Solving for the time period $T$:
$T^2=\frac{4{\pi }^{2}R}{g}$
$T=2\pi \sqrt{\frac{R}{g}}$

Given in the problem:
Radius of the circle, $R=4\text{m}$
Acceleration due to gravity, $g\approx {\pi }^2{\text{m/s}}^2$ (or $g\approx 9.8{\text{m/s}}^2$, which is very close to $9.87\approx {\pi }^2$)

Substituting these values into the formula:
$T=2\pi \sqrt{\frac{4}{{\pi }^2}}$
$T=2\pi \cdot \frac{2}{\pi }$
$T=4\text{sec}$

Therefore, the time period of revolution is 4 seconds.