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  • In a resonance pipe, the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. what will be the end correction?Option: 1

In a resonance pipe, the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. what will be the end correction?

Option: 1

1.05 cm


Option: 2

1115.5 cm


Option: 3

92.5 cm


Option: 4

113 cm


Answers (1)

best_answer

For end correction 

\mathrm{\begin{aligned} & x, \frac{l_2+x}{l_1+x}=\frac{3 \lambda / 4}{\lambda / 4} \\ & =3 \\ & x=\frac{l_2-3 l_1}{2} \\ & x=\frac{70.2-3 \times 22.7}{2} \\ & x=1.05 \mathrm{~cm} \end{aligned}}

 

Posted by

Divya Prakash Singh

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