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MHT CET202522 Apr 2025Evening ShiftPhysicsRotational MotionActual

A thin metal wire of length ' L ' and mass ' M ' is bent to form semicircular ring as shown. The moment of inertia about XX ^1 is

Options

  1. AM L^2 4 ^2
  2. B2 ML ^2 ^2
  3. CM L^2 2 ^2
  4. DML ^2 ^2

Correct answer

C. M L^2 2 ^2

Step-by-step solution

The moment of inertia about a diameter of a semicircular ring of mass M and length L requires relating the geometric parameters. For a semicircle of radius R , the arc length gives L = R , so R = L / . The moment of inertia about diameter XX' is derived by integration. A mass element dm = (M / ) d at angle has perpendicular distance R from the axis, contributing dI = dm (R )^2 = (M R^2 / ) ^2 d . Integrating from 0 to : I = M R^2 ₀^ ^2 d = M R^2 ₀^ 1 - 2 2 d = M R^2 2 [ - 2 2 ]₀^ = M R^2 2 Substituting R = L / yiel

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