If frequency of organ pipe is independent of temperature, then value of `alpha` (coefficient of linear expansion of material of pipe) should be nearly

Options

A. `(1)/(273)// ^(@)C`
B. `(1)/(546)// ^(@)C`
C. `(1)/(819)// ^(@)C`
D. `(1)/(1092)// ^(@)C`

Answer

B

Explanation

`f prop (v)/(t)`
`(f_(t^(@)C))/(f _(0^(@)C)) = (v_(t))/(v_(0)).(l_(0))/(l_(t)) = sqrt((273+t)/(273)) (l_(0))/(l_(0)(1+ alpha t))`
`= ((1+t//273)^(1//2))/(.(1+ alpha t)) = ((1+t//546))/((1+ alpha t))`
If frequency is independent of temerature
`alpha =(1)/(546)//^(@)C`