69.
If I1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass, and I2 is the moment of inertia of the ring formed by bending the rod, then
(a) I1 : I2 = 1 : 1
(b) I1 : I2 = p2 : 3
(c) I1 : I2 = p : 4
(d) I1 : I2 = 3 : 5
Ans. b
(4)
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(a) – 5 J (b) – 10 J (c) – 15 J (d) – 20 J Answer-a (123)