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(D) To prevent water from spilling at the highest point of the vertical circle,the centripetal force must be at least equal to the gravitational force

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) To prevent water from spilling at the highest point of the vertical circle,the centripetal force must be at least equal to the gravitational force.At the highest point,the condition for the water not to spill is $mg = m\omega^2 r$,where $r = 4 \ m$ is the radius (length of the string) and $\omega$ is the angular velocity.Using $\omega = \frac{2\pi}{T}$,we get $g = \left(\frac{2\pi}{T}ight)^2 r$.Substituting the values ($g = 10 \ m/s^2$ and $r = 4 \ m$):$10 = \frac{4\pi^2}{T^2} 	imes 4$.Assuming $\pi^2 \approx 10$,we get $10 = \frac{4 	imes 10 	imes 4}{T^2}$.$10 = \frac{160}{T^2} \implies T^2 = 16$.Therefore,$T = 4 \ s$.

$A$ bucket filled with water is tied to a string of length $4 \ m$ and rotated in a vertical circle. To ensure that the water does not spill,the time period of rotation should be ........ $s$.

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