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  • Tell me? In a resonance pipe, the first and 2nd resonance are obtained at depts 22.7 cm & 70.2 cm. What will be end correction

In a resonance pipe, the first and 2nd resonance are obtained at depts 22.7 cm & 70.2 cm. What will be end correction

  • Option 1)

    1.05 cm

  • Option 2)

    1115.5 cm

  • Option 3)

    92.5 cm

  • Option 4)

    113.5 cm

 

Answers (1)

best_answer

\frac{l_{2}+e}{l_{1}+e}= \frac{3\lambda /4}{\lambda /4}= 3

e= \frac{l_{2}-3l_{1}}{2}= \frac{70.2-3\times 22.7}{2}

e= 1.05cm

 

Fundamental frequency with end correction -

\nu _{0}= \frac{V}{4\left ( l+e \right )}     (one end open)

\nu _{0}= \frac{V}{2\left ( l+2e \right )}    (Both end open)

e = end correction

-

 

 


Option 1)

1.05 cm

This is correct

Option 2)

1115.5 cm

This is incorrect

Option 3)

92.5 cm

This is incorrect

Option 4)

113.5 cm

This is incorrect

Posted by

divya.saini

View full answer

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