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STD 11 - 14. waves and sound — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 14. waves and sound4 Marks
MCQ
A string is fixed at both ends vibrates in a resonant mode with a separation $2.0 \,\,cm$ between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to $1.6\,\, cm$. The length of the string is .... $cm$
  • A
    $4$
  • ✓
    $8$
  • C
    $12$
  • D
    $16$
✓

Answer

Correct option: B.
$8$
b
Let there be $n$ loops in the $1\, st$ case

$\rightarrow$ Length of the wire$L$ $=\left(\left(\mathrm{n} \lambda_{1}\right) / 2\right)$

$\left(\lambda_{1}=2 \times 2=4 \mathrm{cm}\right)$

$\rightarrow$ Length of the wire $L_{l}=\left\{(n+1) \frac{\lambda_{2}}{2}\right\}$

$\left(\lambda_{2}=2 \times(1.6)=3.2 \mathrm{cm}\right)$

$\rightarrow \frac{n \lambda_{1}}{2}=(n+1) \frac{\lambda_{2}}{2}$

$\rightarrow n \times 4=(n+1)(3.2)$

$\rightarrow 4 n-(3.2) n=3.2$

$\rightarrow 0.8 n=3.2$

$\rightarrow n=4$

Length of the string

$L=\frac{n \lambda_{1}}{2}=\frac{(4 \times 4)}{2}=8 \mathrm{cm}$

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