Q112 / 150 · MHT CET 2025 (5 May Shift 2) · Physics
A stone of mass ' m ' kg is tied to a string of length ' L ' m and moved in a vertical circle of radius 49 cm in a vertical plane. If it completes 30 revolutions per minute, the tension in the string when it is at the lowermost point is nearly
[Take $\pi^2=10$ and acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
- A$(90 \mathrm{~m}) \mathrm{N}$
- B$(60 \mathrm{~m}) \mathrm{N}$
- C$(45 \mathrm{~m}) \mathrm{N}$
- Correct answerD$(15 \mathrm{~m}) \mathrm{N}$
Solution
The mass is $m$ kg, with string length (radius) $L = 0.49\text{ m}$, and rotational frequency $f = 0.5\text{ rps}$. Angular velocity is $\omega = 2\pi f = \pi\text{ rad/s}$.
Centripetal acceleration becomes $a_c = \omega^2 L = (\pi)^2 (0.49) = 10 \times 0.49 = 4.9\text{ m/s}^2$.
At the lowermost point, the net centripetal force satisfies $T - mg = ma_c$. Solving yields the tension: $T = m(g + a_c) = m(10 + 4.9) = 14.9m\text{ N}$.
This gives $T \approx 15m\text{ N}$, corresponding to option D.
Hence, the correct answer is option: 'D'.