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A steel wire of length 1 m and mass 0.1 ...

A steel wire of length `1 m` and mass `0.1 kg` and having a uniform cross-sectional area of `10^(-6) m^(2)` is rigidly fixed at both ends. The temperature of the wire is lowered by `20^(@)C`. If the wire is vibrating in fundamental mode, find the frequency (in Hz). `(Y_(steel)=2xx 10^(11) N//m^(2),alpha_(steel)=1.21 xx 10^(-5)//.^(@)C)`.

A

11

B

20

C

15

D

10

Text Solution

Verified by Experts

The correct Answer is:
A

`Delta l=alpha l Delta theta rArr Y=(.^(T)//_(A))/(.^(Deltal)//_(l))rArr T=YA(Deltal)/(l)rArr T=alpha YA Delta theta =48.4N ,v=sqrt((T)/(mu))=sqrt((48.4)/(((0.1)/(1))))=22m//s`
`:.` for fundamental note `l=(lamda)/(2) rArr lamda =2m rArr f=(v)/(lamda)=(22)/(2)=11Hz`.
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