Question

A disc has mass 'M' and radius 'R'. The axis of rotation passes through a point P in the plane of the disc, which is at a distance 'x' \((x\lt R)\) from the center of the disc and perpendicular to the plane of the disc. M.I. of disc about the axis of rotation is found to be \(MR^{2}\). The value of 'x' is

• A. \(2R\)
• B. \(\sqrt{2}\ R\)
• C. \(\frac{R}{\sqrt{2}}\)
• D. \(\frac{R}{2}\)

Hint:

Apply the parallel axis theorem with the central value half of MR squared.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
About the central axis perpendicular to the plane, \(I_0=\dfrac{1}{2}MR^{2}\). The axis through P is parallel to it, so the parallel axis theorem applies.

Step 2: Calculation
\(I=\dfrac{1}{2}MR^{2}+Mx^{2}\)
Given \(I=MR^{2}\).

Step 3: Solve for x
\(Mx^{2}=\dfrac{1}{2}MR^{2}\), so \(x^{2}=\dfrac{R^{2}}{2}\) and \(x=\dfrac{R}{\sqrt{2}}\).

Step 4: Check
This is less than \(R\), as the question requires.

Final Answer:
The distance is R divided by root 2. \[ \boxed{\text{(C) }\dfrac{R}{\sqrt{2}}} \]