A circular disc of radius b has a hole of radius a at its centre(see figure). If the mass per unit area of the disc varies as σ 0 r then, the radius of gyration of the disc about its axis passing through the center is

Step-by-step solution

Solution: (B) d m = σ     2 π r d r = σ 0 r 2   π r d r m =2 π σ 0 ∫ a b d r = 2 π   σ 0 ( b - a ) I = 2 π   σ 0 ∫ a b r 2 d r = 2 3 π σ 0 b 3 - a 3 Radius of gyration = I m   = 2 3 π   σ 0 b 3 - a 3 2 π   σ 0 b - a = b 3 - a 3 3 ( b - a ) = ( b - a ) ( b 2 + a 2 + a b ) 3    ( b - a ) = b 2 + a 2 + a b 3 .