Question-36 Two circular loops P and Q of radii r and nr are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is 4 times that of loop P about its axis. The value of n is
(A) (2)^{-2/3}
(B) (2)^{2/3}
(C) \sqrt{2}
(D)(2) ^{1/3}
Answer : (B)
Explanation:
The two loops are formed from the same wire.
Hence, the linear density will remain constant.
∴ λ = \frac{M_P}{L_P} = \frac{M_Q}{L_Q}
∴ M_P = λ × L_P = λ × (2πR_P) = 2πrλ
Similarly,
M_Q = λ × L_Q = λ × (2πR_Q) = 2π(nr)λ
Also, M.I. of circular loop P is,
I_P = M_P r^2 = 2πrλ(r)^2 = 2πr^3λ
and that of loop Q,
I_Q = M_Q (nr)^2 = 2π(nr)λ × (nr)^2 = 2πn^3r^3λ
Given that, I_Q = 4I_P
∴ 2πn^3r^3λ = 4 × 2πr^3λ
∴ n^3 = 4
∴ n = ^{\frac{1}{3}} = (2)^{\frac{2}{3}}