When the length of the vibrating segment of a sonometer wire is increased by `1%` the percentage change in its frequency is

Options

A. `100/101`
B. `99/100`
C. 1
D. 2

Answer

c

Explanation

The frequency of sonometer wire is
`n=1/(2l)sqrt(T/m)`
Taking logarithm and differentiat, we get
`(Deltan)/n= (Deltal)/l+1/2 (DeltaT)/T+1/2 (Deltam)/m`
`therefore (Deltan)/n= (Deltal)/l+0+0=1%`
Hence, frequency will increases by `1%`.