Question

in the resonance tube experiment 3 rd experiment occurs at 111 and eleven cents and 5 th resonance occurs at 200 find the correction no spam please
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Answer 1

Answer:

The end  correction  is 0.25 cm

Explanation:

The end correction is represented with ( e) .

$l_{1}$ = length of air column at first resonance and

$l_{2}$ is length of air column at second resonance.

λ is wavelength  

Resonance of tube of 3 rd experiment occurs at 111 cm and

5 th resonance occurs at 200 cm.

let us consider from the given data

     $\frac{5}{4}$ λ=$l_{1} +e........(1)$

     $\frac{9}{4}$ λ=$l_{2} +e........(2)$

From (1)and (2) equations

   λ=$l_{2} -l_{1}$

   λ=200-111

  λ=89  cm

  End correction is (e)=$\frac{5}{4}$ λ-$l_{1}$

                                (e)=$\frac{5}{4} *89-111$

                            (e)= 0.25cm

The end  correction  is 0.25 cm

 

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