Q139 / 180 · KCET 2011 · Physics
A wire under tension vibrates with a fundamental frequency of $600 \mathrm{~Hz}$. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become
- Correct answerA$200 \mathrm{~Hz}$
- B$300 \mathrm{~Hz}$
- C$600 \mathrm{~Hz}$
- D$400 \mathrm{~Hz}$
Solution
Frequency of wire, $f=\frac{1}{2 l r} \sqrt{\frac{T}{\pi \rho}}$
So here,
$$
\begin{aligned}
\frac{f_{1}}{f_{2}} &=\frac{l_{2}}{l_{1}} \times \frac{r_{2}}{r_{1}} \times \sqrt{\frac{T_{1}}{T_{2}}} \\
\frac{600}{f_{2}} &=\frac{2}{1} \times \frac{1}{2} \times \sqrt{\frac{T}{T / 9}} \\
\frac{600}{f_{2}} &=3 \\
f_{2} &=\frac{600}{3} \\
f_{2} &=200 \mathrm{~Hz}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.