A stone of mass $$m$$, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

A stone of mass $$m is whirled in a vertical circle of radius r with uniform speed v$$. At any point in the circle, the net radial force provides the centripetal acceleration. Let $$\theta$$ be the angle measured from the top of the circle.

At the highest point of the circle, both weight $$mg and tension T$$ act toward the center (downward is toward the center): $$T_{\text{top}} + mg = \frac{mv^2}{r} leading to T_{\text{top}} = \frac{mv^2}{r} - mg$$.

At the lowest point of the circle, tension acts upward (toward the center) and weight acts downward (away from the center): $$T_{\text{bottom}} - mg = \frac{mv^2}{r} giving T_{\text{bottom}} = \frac{mv^2}{r} + mg$$.

At a horizontal position, the weight has no radial component, so $$T_{\text{horizontal}} = \frac{mv^2}{r}$$.

Comparing these tensions, we have $$T_{\text{top}} = \frac{mv^2}{r} - mg\quad(\text{smallest}),\quad T_{\text{horizontal}} = \frac{mv^2}{r}\quad(\text{intermediate}),\quad T_{\text{bottom}} = \frac{mv^2}{r} + mg\quad(\text{largest}). Since the speed is uniform, the tension is minimum at the highest position of the circular path, where gravity assists in providing the centripetal force.

The correct answer is Option C.