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 'm' are attached at distance` 'L//2' ` from its centre on both sides, it reduces the oscillation frequency by `20%`. The value of ratio `m//M` is close to :

Options

A. 0.57
B. 0.37
C. 0.77
D. 0.17

Answer

B

Explanation

`tau Ctheta implies (I(d^(2)theta))/(dt^(2))+Ctheta=`
Where C is tensional constant
`omega_(1)=sqrt(C/I_(1))`
`omega_(2)=sqrt(C/I_(2))`
`omega_(2)=80/100omega_(1)`
`impliesomegaunderset(2)overset(2)`=`0.64omega underset(1)overset(2)` `implies I=0.64I_(2)`
`(M(2L)^(2))/12 =64/100xx[(M4L^(2))/12+(3mL^(2))/(4xx3)xx2]`
`4M=64/100(4M+6M) implies m/M=3/8equiv0.37`