In a resonance tube experiment , the first resonance is obtained for `10 cm` of air column and the sound for `32 cm`. The end correction for this apparatus is

Options

A. `0.5 cm`
B. `1.0 cm`
C. `1.5 cm`
D. `2 cm`

Answer

B

Explanation

` l_(1) + e = lambda//4 or 3 l_(1) + 3 e = 3 lambda //4`
Again `l_(2) + e = ( 3lambda)/( 4)`
`:. 3 l_(1) + 3 e = l_(2) + e`
or ` 2 e = l_(2) - 3 l_(1)`
or ` e = (1)/(2) (l_(2) - 3 l_(1)) = (1)/(2) ( 32 - 3 xx 10) = 1 cm`