Question

A circular disc of radius R is removed from a ** disc of radius 2R. such that the circumferences of the disc coincide. The centre of mass of the remaining portion is αR from the centre of mass of the ** disc. The value of α is.

(A) 1/2

(B) 1/6

(C) 1/4

(D) [(– 1) / 3]

Answer 1

Correct option: (D) [(– 1) / 3]

Explanation:

Let m be mass per unit area

M = mass of big disc = π(2R)2m

M1 = mass of removed disc = πR2m

M2 = M – M1 = Mass of remaining Portion

M2 = M – M1 = Mass of remaining Portion

M× O = M1(OO') + (M – M1)(OG)

M1 ∙ R = – (M – M1) (∝R)

M1 = – α(M – M1)

πR2m = – α(π ∙ 4R2m – πR2m)

R2 = – α(3R2)

α = [(– 1) / 3]