Correct Answer - A
For string fixed at both the ends, resnant frequency are given by `f = (nv)/(2 L)`. It is given that `315 Hz` and `420 Hz` are two consecutive resonant frequencies. Let these be `nth` and `(n + 1) th` harmonices.
`315 = (nv)/(2 L)` ..(i)
`420 = ((n + 1)v)/(2 L)`...(ii)
Dividing Eq. (i) by Eq. (ii), we get
`(315)/(420) = (n)/( n + 1) rArr n = 3`
Lowest resonant frequency,
`f_0 = (v)/(2 L) = (315)/(3) = 105 Hz`.