Question

A bucket containing water is revolved in vertical circle of radius r. To prevent the water from falling down, the period of revolution required is (\(g\) = gravitational acceleration)

• A. \(2π\sqrt{\frac{g}{r}}\)
• B. \(2π\sqrt{\frac{r}{g}}\)
• C. \(2π\sqrt{rg}\)
• D. \(\frac{\sqrt{rg}}{2π}\)

Hint:

At the top, minimum speed satisfies mg = mv^2/r.

Answer 1

The Correct Option is B

Step 1: Top condition
$mg \leq mv^2/r$, with equality at the limit, so $v_{\min}^2 = rg$.

Step 2: Angular speed
$\omega = v/r = \sqrt{g/r}$.

Step 3: Period
$T = 2\pi/\omega = 2\pi\sqrt{r/g}$. Option (B).

Final Answer:
Option (B). \[ \boxed{2\pi\sqrt{\frac{r}{g}}} \]