Question

A metallic wire with tension T and at temperature ${30}^{\circ }$C vibrates with its fundamental frequency of 1kHz. The same wire with the same tension but at ${10}^{\circ }$C temperature vibrates with a fundamental frequency of 1.001 kHz. The coefficient of linear expansion of the wire is

• A. $2\times {10}^{-4}{/}^{\circ }$C
• B. $1.5\times {10}^{-4}{/}^{\circ }$C
• C. $1\times {10}^{-4}{/}^{\circ }$C
• D. $0.5\times {10}^{-4}{/}^{\circ }$C

Answer 1

Answer: (C)

$1\times {10}^{-4}{/}^{\circ }$C

$f_{30}=1$kHz and $f_{10}=1.001$kHz

We know,

$l_2=l_1(1+\alpha dt)$, where dt $=$ change in temperature

Now, $f=\frac{1}{2\pi }\sqrt{\frac{g}{l}}$

So, $f\alpha \frac{1}{\sqrt{l}}$

$\frac{f_{30}}{f_{10}}=\sqrt{\frac{l_{10}}{l_{30}}}=\frac{1}{1.001}$

$\frac{l_{10}}{l_{30}}=\frac{1}{(1.001{)}^2}=\frac{1}{(1+\alpha dt)}=\frac{1}{(1+20\alpha )}$

Solving we get, $\alpha =1\times {10}^{-4}{/}^{\circ }$C