MCQ
In the resonance tube experiment, first resonant length is $l_1$ and second resonant length is $l_2$ then the third resonant length will be
- A$5l_2$
- B$2(l_2-l_1)$
- ✓$2l_2-l_1$
- D$3l_2-2l_1$
✓
Answer
Correct option: C.
$2l_2-l_1$
c
for first resonance
for first resonance
$\ell_{1}+\varepsilon=\frac{v}{4 \mathrm{f}_{0}}$
for second resonance
$\ell_{2}+\varepsilon=\frac{3 \mathrm{v}}{4 \mathrm{f}_{0}}$
$\ell_{3}+\varepsilon=\frac{5 \mathrm{v}}{4 \mathrm{f}_{0}}$
Solving get $\ell_{3}=2 \ell_{2}-\ell_{1}$
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