Question

A sphere of mass \( M \) is attached to one end of a metal wire having length \( L \) and diameter \( D \). It is whirled in a vertical circle of radius \( R \) with angular velocity \( \omega \). When the sphere is at the lowest point of its path, the elongation of the wire is

• A. \( \frac{6ML \left( R^2 \omega^2 + g \right)}{\pi D^2 Y} \)
• B. \( \frac{2ML \left( R^2 \omega^2 + g \right)}{\pi D^2 Y} \)
• C. \( \frac{4ML \left( R \omega^2 + g \right)}{\pi D^2 Y} \)
• D. \( \frac{ML \left( R \omega^2 + g \right)}{2 D^2 Y} \)

Hint:

For elongation due to a force, use Hooke's Law \( \Delta L = \frac{F L}{A Y} \), and remember to calculate the force acting on the object.

Answer 1

The Correct Option is C

Step 1: Understanding the elongation of the wire.
The elongation \( \Delta L \) of a wire due to a force is given by Hooke's Law, which is: \[ \Delta L = \frac{F L}{A Y} \] where \( F \) is the force applied, \( L \) is the length of the wire, \( A \) is the cross-sectional area, and \( Y \) is Young’s modulus. The force acting on the sphere at the lowest point of its path is the centripetal force and the gravitational force. This force is given by: \[ F = M \left( R \omega^2 + g \right) \] where \( \omega \) is the angular velocity and \( g \) is the acceleration due to gravity. Step 2: Calculating the cross-sectional area.
The cross-sectional area of the wire is \( A = \frac{\pi D^2}{4} \), where \( D \) is the diameter of the wire. Step 3: Substituting into the formula for elongation.
The elongation of the wire is then: \[ \Delta L = \frac{F L}{A Y} = \frac{M L \left( R \omega^2 + g \right)}{\pi D^2 Y} \cdot 4 \] Thus, the correct answer is \( \frac{4ML \left( R \omega^2 + g \right)}{\pi D^2 Y} \), which corresponds to option (C).