Question

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 (g = 10 m/s^2)]

• A. ((90 m) N)
• B. ((60 m) N)
• C. ((45 m) N)
• D. ((15 m) N)

Hint:

Tension is maximum at the bottom ($mg + mr\omega^2$) and minimum at the top ($mr\omega^2 - mg$).

Answer 1

The Correct Option is D

Step 1: Parameters
$r = 0.49 \text{ m}$.
$\omega = 30 \text{ rpm} = \frac{30 \times 2\pi}{60} = \pi \text{ rad/s}$.

Step 2: Tension Formula at Bottom
$T_L = mg + mr\omega^2$.

Step 3: Calculation
$T_L = m(10) + m(0.49)(\pi^2)$.
Using $\pi^2 = 10$: $T_L = 10m + 0.49(10)m = 10m + 4.9m = 14.9m \text{ N}$.
Nearly $15m \text{ N}$.
Final Answer: (D)