Q149 / 150 · MHT CET 2022 (07 Aug Shift 1) · Physics
A uniform wire of length $L$, diameter $D$ and density $\rho$ is stretched by a tension $T$ length $L$ and the diameter $D$ is
- A$f \propto \frac{L}{D}$
- Correct answerB$f \propto \frac{1}{L D}$
- C$f \propto \frac{1}{L \sqrt{D}}$
- D$f \propto \frac{1}{L D^2}$
Solution
Given,
Frequency $=f$, Diameter $=D$, Length $=L$, Density $=\rho$ and Tension $=T$
Now, we know that $f=\frac{c}{\lambda}$,
$\therefore f=\frac{1}{2 L} \times \sqrt{\frac{T}{\mu}}$
where $\mu$ is mass per unit length
And mass per unit length is related to density via,
$\mu=\left(\pi \times \frac{D^2}{4}\right) \frac{1}{\rho}$
So, the frequency is $f=\frac{1}{2 L} \times \sqrt{\frac{T}{\left(\frac{\pi D^2}{4 \rho}\right)}}$
$\therefore f \propto \frac{1}{L D}$
Hence, the correct answer is option: 'B'.