Question

A stone of mass \(2\,\text{kg}\) attached at one end of a \(2\,\text{m}\) long string is whirled in a horizontal circle. The string makes an angle of \(45^\circ\) with the vertical. Then the centripetal force acting on the stone is (\(g = 10\,\text{m s}^{-2}\), \(\tan 45^\circ = 1\))

• A. \(30\ \text{N}\)
• B. \(40\ \text{N}\)
• C. \(20\ \text{N}\)
• D. \(10\ \text{N}\)

Hint:

In a conical pendulum, centripetal force is always given by \(mg\tan\theta\).

Answer 1

The Correct Option is C

Step 1: Identify forces acting on the stone.
The stone experiences tension \(T\) in the string and its weight \(mg\) acting vertically downward.

Step 2: Resolve tension into components.
Vertical component balances weight:
\[ T\cos\theta = mg \] \[ T = \frac{mg}{\cos 45^\circ} \]
Step 3: Determine centripetal force.
The horizontal component of tension provides the centripetal force:
\[ F_c = T\sin\theta \] \[ F_c = mg\tan\theta \]
Step 4: Substitute values.
\[ F_c = 2 \times 10 \times 1 = 20\ \text{N} \]
Step 5: Conclusion.
The centripetal force acting on the stone is \(20\ \text{N}\).