Q.
In a resonance tube experiment, 3rd resonance occurs at 111cm and 5th resonance occurs at 200cm . Find the end correction (in cm ) is α . Write the value of 100α .
In the resonance tube, for 3rd resonance, 45λ=l1+e .....(i)
For 5th resonance, 49λ=l2+e .....(ii)
From (i) and (ii), λ=l2−l1=200−111=89cm
From (i), end correction e=45λ−l1=45×89−111 ⇒e=0.25cm=α100α=25