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Fundamental frequency of a stretched son...

Fundamental frequency of a stretched sonometer wire is `f_(0)`. When its tension is increased by `96%` and length drecreased by `35%`, its fundamental frequency becomes `eta_(1)f_(0)`. When its tension is decreased by `36%` and its length is increased by `30%`, its fundamental frequency becomes `eta_(2)f_(0)`. Then `(eta_(1))/(eta_(2))` is.

A

`4.5`

B

`3.5`

C

`2`

D

`6`

Text Solution

Verified by Experts

The correct Answer is:
B
`f_(0) = (1)/(2L) sqrt((T)/(mu))`
`f = (1)/(2(L + DeltaL))sqrt((T + DeltaT)/(mu)) = etaf_(0)`
`:. eta = (L)/(L + DeltaL)sqrt((T + DeltaT)/(T)) = sqrt(1 + (DeltaT)/(T))/(1 + (DeltaT)/(T))`
`:. eta_(1) = (sqrt(1 - 0.96))/(1 - 0.35) = (1.4)/(0.65)` and `eta_(2) = (sqrt(1-0.36))/(1+0.3) = (0.8)/(1.3)`
`:. (eta_(1))/(eta_(2)) = (1.4)/(0.65) xx (1.3)/(0.8) = 3.5`
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