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14. waves and sound — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics14. waves and sound1 Mark
MCQ
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 \mathrm{~cm}$, the length of the closed pipe will be :
  • A
    $60 \mathrm{~cm}$
  • B
    $45 \mathrm{~cm}$
  • C
    $30 \mathrm{~cm}$
  • ✓
    $15 \mathrm{~cm}$
✓

Answer

Correct option: D.
$15 \mathrm{~cm}$
d
$\frac{\lambda}{4}=\mathrm{L}_1 \quad 2\left(\frac{\lambda}{2}\right)=\lambda$

$\mathrm{v}=\mathrm{f} \lambda \quad \mathrm{f}_2=\frac{2 \mathrm{v}}{2 \mathrm{~L}_2}$

$\mathrm{v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right) \quad \mathrm{f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}$

${v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right)$

${f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}$

${f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1}$

${f}_1=\mathrm{f}_2$

$\frac{\mathrm{V}}{4 \mathrm{~L}_1}=\frac{\mathrm{V}}{\mathrm{L}_2}$

$\Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1$

$60=4 \times \mathrm{L}_1$

${~L}_1=15 \mathrm{~cm} $

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