Question

A stone of mass \(40\) g is tied to a light inextensible string of length \(50\) cm and whirled in a vertical circle at the rate of \(30\) revolutions per minute. The tension in the string when it is at the lowest point of the circle is
(Take gravitational acceleration, \(g = 10\,\text{m/s}^2\) and \(π^2 = 10\))

• A. \(0.4\) N
• B. \(0.5\) N
• C. \(0.6\) N
• D. \(0.7\) N

Hint:

At the lowest point the tension must supply the weight plus the centripetal force.

Answer 1

The Correct Option is C

Step 1: Understand the concept
At the lowest point both the weight (down) and the tension (up) act. The net upward force provides the centripetal force: \(T - mg = m\omega^2 r\).

Step 2: Convert the data
\(m = 40\ \text{g} = 0.04\) kg, \(r = 50\ \text{cm} = 0.5\) m. The frequency is \(30\ \text{rev/min} = 0.5\) rev/s, so \(\omega = 2\pi f = \pi\) rad/s.

Step 3: Compute
\[ m\omega^2 r = 0.04 \times \pi^2 \times 0.5 = 0.04 \times 10 \times 0.5 = 0.2\ \text{N} \]
\[ mg = 0.04\times10 = 0.4\ \text{N} \]

Step 4: Result
\(T = 0.4 + 0.2 = 0.6\) N, option (C). The value 0.4 N is the weight alone and 0.2 N is the centripetal term alone.

Final Answer:
The tension is 0.6 N. This is option (C). \[ \boxed{\text{(C) }0.6\ \text{N}} \]