A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations , If two masses each of mass ‘m’ are attached at distance $\frac{L}{2}$ from centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is closed to

detailed solution

Correct option is B

I L lmoment of inertia of rod is, I=M2L212, here M is mass of rod,2L is length of rodmoment of inertia of rod with masses m attached is,                              I1=I+2mL22---(1) now,  I=ML23---(2)  on substituting equation 2 in 1, I1                                             =L2M3+m2                                I1=L22M+3m6frequency of torsional oscillations is f=12πKI---(3)     ⇒                                                               f1=12πkI1---(4)  So by question,f−f1f=1−f1f=20100=15  on substituting equations 3 and 41−f1f=1−II1 ⇒15=1−II1 ⇒1−15=II1 ⇒452=II1  substitute the values of I,I1 ⇒ML23L22M+3m6=16252M2M+3m=162525M=16M+24m9M=24m∴ mM=38=0.37