MHT CET · Physics · Waves and Sound
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
- A decreased by \(10 \%\)
- B decreased by \(20 \%\)
- C decreased by \(5 \%\)
- D decreased by \(15 \%\)
Answer & Solution
Correct Answer
(B) decreased by \(20 \%\)
Step-by-step Solution
Detailed explanation
(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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