A stone of mass 0.3 kg tied to the end of a string is whirled round in a circle of radius 1.2 m with a speed of 20 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 100 N?

Correct option is D

​$$\text{Given:} \\\text{Mass of the stone, } m = 0.3 \, \text{kg} \\\text{Radius of the circle, } r = 1.2 \, \text{m} \\\text{Speed of the stone, } v = 20 \, \text{rev/min} \\\text{Concept:} \\\text{the centrifugal force is balanced by the tension force (centripetal force) when the stone is whirled in a circle.} \\\text{The formula for the tension } T \text{ is:} \\T = m \cdot r \cdot \omega^2 \\\text{Where:} \\T = \text{Tension in the string,} \\m = \text{Mass of the stone,} \\r = \text{Radius of the circle,} \\\omega = \text{Angular velocity.} \\\text{The relationship between speed } v \text{ and angular velocity } \omega \text{ is:} \\\omega = 2\pi \times \left( \dfrac{\text{rev/min}}{60} \right)$$

$$\omega = 20 \, \text{rev/min} = 20 \times 2 \times \left( \dfrac{\pi}{60} \right) = 2.094 \, \text{rad/s} \\\text{ the tension in the string:} \\T = m \cdot r \cdot \omega^2 \\T = 0.3 \times 1.2 \times (2.094)^2 \\T = 1.57 \, \text{N} \\\text{the maximum speed when the tension is 100 N:} \\\text{ } T = m \cdot r \cdot \omega^2, \text{ and substituting } T = 100 \, \text{N:} \\100 = 0.3 \times 1.2 \times \omega^2 \\\omega^2 = \dfrac{100}{0.3 \times 1.2} = 277 \\\omega = \sqrt{277} = 16.167 \, \text{rad/s} \\\text{Now, calculate the maximum speed } v \text{ using the relationship } v = r \cdot \omega: \\v = 1.2 \times 16.167 = 20 \, \text{m/s}$$​​​