A stone of mass ' m ' kg is tied to a string of length ' L ' m and moved in a vertical circle of radius 49…
- $(90 \mathrm{~m}) \mathrm{N}$
- $(60 \mathrm{~m}) \mathrm{N}$
- $(45 \mathrm{~m}) \mathrm{N}$
- $(15 \mathrm{~m}) \mathrm{N}$
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.
Asked in: MHT CET 2025 (05 May Shift 2)