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In resonance tube exp we find l(1) = 25....

In resonance tube exp we find `l_(1) = 25.0 cm` and `l_(2) = 75.0cm` The least count of the scale used to measure l is `0.1cm` If there is no error in frequency What will be max permissible error in speed of sound (take `f_(0) = 325 Hz)` .

Text Solution

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`V = 2f_(0) (l_(2) - l_(1))`
`(dV) = 2f_(0) (dl_(2) - dl_(1))`
`(DeltaV)_(max) = max of [2f_(0) (+- Deltal_(2) + Deltal_(1))`
`Deltal_(1)=` least count of the scale `= 0.1 cm`
`Deltal_(2)=` least count of scale `= 0.1 cm`
So max permissible error in speed of sound `(DeltaV)_(max) = 2 (325Hz) (0.1 cm + 0.1 cm) = 1.3 m//s`
Value of `V = 2f_(0) (l_(2) -l_(1)) = 2 (325 hz) 975.0cm - 25.0 cm) = 325 m//s`
so `V = (325 +- 1.3)m//s` .
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