A rod of mass M and length 2 L is performing SHM as torsional pendulum in the horizontal plane. Two blocks, each of mass m , are put at distance L 2 from the centre. The frequency after putting blocks of mass m is 20 % of initial frequency. Then, the ratio of m M will be

Step-by-step solution

T = 2 π   I C f = 1 2 π   C I f i n i t i a l = 1 2 π   3 C M L 2 f f i n a l = 1 2 π C M L 2 3 + 2 m L 2 4 0.2   1 2 π   3 C M L 2 = 1 2 π C M L 2 3 + 2 M L 2 4   = 0.04 3 M = 12 4 M + 6 M 4 M = 0.16 M + 0.24   M m M = 3.84 0.24 = 16