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)` .

Explanation

`V = 2f_(0)(l_(2) - l_(1))`
`(dV) = 2f_(0) ( dl_(2) - dl_(1))`
`(dV)_(max) = max of [ 2f_(0) (+- Delta l_(2) +- Deltal_(2)]`
`l_(1) = 25.0 cm rArr l_(1) = 0.1 cm` ( place value of last number)
`l_(1) = 75.0 cm rArr l_(2) = 0.1 cm` ( place value of last number)
So the maximum permissible error in the speed of sound `(dV)_(max)`
`= 2( 325 Hz) ( 0.1 cm + 0.1 cm ) = 1.3 m s^(-1)`
Value of `V = 2f_(0) (l_(2) - l_(1) = 2(325 Hz) (75.0 cm - 25.0 cm)`
` = 325 m s^(-1)`
So `V = (325 +- 1.3) m s^(-1)`.