The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ___________ m.
Given, the length of the rod is 10 √3 m.

Solution

Physics sir · 4 steps

Write I for a thin rod about its centre, set it equal to Mk², and the mass drops out.

1. Moment of inertia. Treat the rod as thin. About an axis through its centre and perpendicular to its length, I = ML²/12.
2. Define k. The radius of gyration k is defined by I = Mk². Equate the two expressions, and M cancels.

   Mk² = ML²/12 ⇒ k = L/√12 = L/2√3

3. Put in numbers. With L = 10√3 m, k = 10√3/2√3 m. The √3 cancels, leaving 10/2 m.
4. So. The radius of gyration is 5 m.

The common slip: 10. Using I = ML²/3, which is for an axis through one end, gives k = L/√3 = 10 m.

The idea to keep. Radius of gyration comes from I = Mk², so k = √(I/M). For a rod about its centre, k = L/√12. Always check where the axis passes before picking the formula for I.

Answer from the NTA final answer key. Solution by PrepWiser.