ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

The moment of inertia of semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is \(\frac{1}{ x } MR ^2\), where \(R\) is the radius and \(M\) is the mass of semicircular ring. The value of \(x\) will be \(...........\)

  1. A \(2\)
  2. B \(1\)
  3. C \(3\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

The moment of inertia of semicircular ring about axis passing through centre of ring and perpendicular to plane of ring is \(= MR ^2\) so \(x=1\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app