JEE Physics 2024 Question with Solution

Given: Mass of the stone $m = \frac{900}{1000} \, \text{kg} = 0.9 \, \text{kg}$, radius $r = 1 \, \text{m}$, speed of rotation $N = 10 \, \text{rpm}$, and gravitational acceleration $g = 9.8 \, \text{m/s}^2$.

Find: The tension $T$ in the string at the lowest point of the vertical circle.

At the lowest point, the centripetal force is upward toward the center, so

$$T - mg = m\omega^2 r$$

Hence,

$$T = mg + m\omega^2 r$$

Now convert revolutions per minute to angular velocity:

$$\omega = \frac{2\pi N}{60}$$ $$\omega = \frac{2 \times 3.14 \times 10}{60} \approx 1.047 \, \text{rad/s}$$

Now calculate the two terms:

$$mg = 0.9 \times 9.8 = 8.82 \, \text{N}$$ $$m\omega^2 r = 0.9 \times (1.047)^2 \times 1 \approx 0.99 \, \text{N}$$

Therefore,

$$T = 8.82 + 0.99 = 9.81 \, \text{N}$$

So the required tension is approximately $9.8 \, \text{N}$. The correct option is B.

Given: $m = 900 \, \text{g} = 0.9 \, \text{kg}$, $r = 1 \, \text{m}$, and rotation rate $10 \, \text{rpm}$.

Find: Tension at the lowest point.

First convert the angular speed:

$$\omega = 10 \times \frac{2\pi}{60} = \frac{\pi}{3} \, \text{rad/s}$$

The centripetal force needed is

$$F_c = m\omega^2 r$$ $$F_c = 0.9 \times \left(\frac{\pi}{3}\right)^2 \times 1 \approx 0.98 \, \text{N}$$

At the lowest point, tension has to balance weight and also provide centripetal force, so

$$T = mg + F_c$$ $$T = (0.9 \times 9.8) + 0.98 = 8.82 + 0.98 = 9.8 \, \text{N}$$

Therefore, the tension in the string at the lowest point is $9.8 \, \text{N}$ and the correct option is B.