Question

A $40cm$ wire having a mass $3.2gm$ and area of cross-section $1m{m}^2$ is stretched between two supports $40.05cm$ apart. In its fundamental note, the wire vibrates with a frequency $220Hz$. The Young's molulus is :

• A. $1.98\times {10}^{11}N/m^2$
• B. $2.2\times {10}^{11}N/m^2$
• C. $3.96\times {10}^{11}N/m^2$
• D. $3.2\times {10}^{11}N/m^2$

Answer 1

The correct option is A $1.98\times {10}^{11}N/m^2$

$l=40cm$

$\Delta l=0.05cm$

$f=\frac{v}{2l}$

$\Rightarrow v=22\times 8=176m/s$

Now$,$ $v=\sqrt{\frac{T}{\mu }}$

$\Rightarrow T=176\times 176\times \frac{3.2\times {10}^{-3}}{40\times {10}^{-2}}$

$=\frac{176\times 176\times 32}{40\times 10\times 10}$

$=\frac{176\times 176\times 32}{4\times {10}^3}=247.8N$

$\therefore \gamma =\frac{247.8\times 40}{1\times {10}^{-6}\times 0.05}$

$=1.98\times {10}^{11}N/m^2$

Hence,

option $\left(A\right)$ is correct answer.