JEE Main 2024 — 29 January, Shift 2 — Question 35

Given: Mass of stone, $m = 900 \text{ g} = 0.9 \text{ kg}$ Radius of vertical circle, $r = 1 \text{ m}$ Rotational speed, $N = 10 \text{ rpm}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$ Value of $\pi^2 = 9.8$

First, convert the rotational speed from rpm to angular velocity ($\omega$) in rad/s: $N = 10 \text{ revolutions/minute}$ $\omega = N \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}}$ $\omega = 10 \times \frac{2\pi}{60} \text{ rad/s}$ $\omega = \frac{20\pi}{60} \text{ rad/s}$ $\omega = \frac{\pi}{3} \text{ rad/s}$

Next, calculate the linear velocity ($v$) of the stone: $v = r\omega$ $v = 1 \text{ m} \times \frac{\pi}{3} \text{ rad/s}$ $v = \frac{\pi}{3} \text{ m/s}$

Now, calculate $v^2$: $v^2 = \left(\frac{\pi}{3}\right)^2 = \frac{\pi^2}{9}$ Given $\pi^2 = 9.8$, so: $v^2 = \frac{9.8}{9} \text{ (m/s)}^2$

At the lowest point of the vertical circle, the tension ($T$) in the string and the weight ($mg$) of the stone act as follows: Weight ($mg$) acts downwards. Tension ($T$) acts upwards. The net force towards the center provides the necessary centripetal force ($F_c = \frac{mv^2}{r}$).

Therefore, the equation for forces at the lowest point is: $T - mg = \frac{mv^2}{r}$ $T = mg + \frac{mv^2}{r}$

Substitute the known values into the equation: $T = (0.9 \text{ kg})(9.8 \text{ m/s}^2) + \frac{(0.9 \text{ kg})\left(\frac{9.8}{9} \text{ m}^2/\text{s}^2\right)}{1 \text{ m}}$ $T = 0.9 \times 9.8 + 0.9 \times \frac{9.8}{9}$ $T = 0.9 \times 9.8 + 0.1 \times 9.8$ Factor out 9.8: $T = 9.8 (0.9 + 0.1)$ $T = 9.8 \times 1$ $T = 9.8 \text{ N}$

The tension in the string when the stone is at the lowest point is $9.8 \text{ N}$.