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The fundamental frequency of a sonometer wire is $50 \mathrm{~Hz}$ for some length and
tension. If the length is increased by $25 \%$ by keeping tension same, then frequency
change of second harmonic is
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tension. If the length is increased by $25 \%$ by keeping tension same, then frequency
change of second harmonic is
Solution:
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The correct answer is:
decreased by $20 \%$
(D)
$\begin{aligned} & \mathrm{n}_{2} \ell_{2}=\mathrm{n}_{1} \ell_{1} \\ \therefore & \mathrm{n}_{2}=\frac{\ell_{1}}{\ell_{2}} \cdot \mathrm{n}_{1}=\frac{\ell_{1}}{1.25 \mathrm{R}_{1}} \mathrm{n}_{1}=0.8 \mathrm{n}_{1} \\ \therefore & \mathrm{n}_{1}-\mathrm{n}_{2}=\mathrm{n}_{1}(1-08)=0.2 \mathrm{n}_{1} \\ & \frac{\mathrm{n}_{1}-\mathrm{n}_{2}}{\mathrm{n}_{1}}=0.2 \end{aligned}$
$\begin{aligned} & \mathrm{n}_{2} \ell_{2}=\mathrm{n}_{1} \ell_{1} \\ \therefore & \mathrm{n}_{2}=\frac{\ell_{1}}{\ell_{2}} \cdot \mathrm{n}_{1}=\frac{\ell_{1}}{1.25 \mathrm{R}_{1}} \mathrm{n}_{1}=0.8 \mathrm{n}_{1} \\ \therefore & \mathrm{n}_{1}-\mathrm{n}_{2}=\mathrm{n}_{1}(1-08)=0.2 \mathrm{n}_{1} \\ & \frac{\mathrm{n}_{1}-\mathrm{n}_{2}}{\mathrm{n}_{1}}=0.2 \end{aligned}$
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