Question

A bucket containing water is revolved in a vertical circle of radius $r$. To prevent the water from falling down, the minimum frequency of revolution required is (where $g$ = acceleration due to gravity)

• A. $\frac{1}{2\pi} \sqrt{\frac{g}{r}}$
• B. $2\pi \sqrt{\frac{g}{r}}$
• C. $\frac{2\pi g}{r}$
• D. $\frac{1}{2\pi} \sqrt{\frac{r}{g}}$

Hint:

In circular motion, the minimum speed at the top of the circle is given by $\sqrt{gr}$, which is derived from balancing forces.

Answer 1

The Correct Option is A

Step 1: Condition for water not falling down.
To prevent the water from falling, the centripetal force must be at least equal to the weight of the water. The condition for the critical velocity is:
\[ v_{\text{min}} = \sqrt{gr} \]
Step 2: Relating velocity and frequency.
The velocity for a body in circular motion is related to frequency $f$ by:
\[ v = 2\pi f r \]
Step 3: Solving for minimum frequency.
Equating the two expressions for $v$:
\[ 2\pi f r = \sqrt{gr} \] \[ f = \frac{1}{2\pi} \sqrt{\frac{g}{r}} \]
Step 4: Conclusion.
The minimum frequency required is $\frac{1}{2\pi} \sqrt{\frac{g}{r}}$.