91Ó°ÊÓ

The given data can be listed below as-

• Length of wire A$l_A=2.8m$
• Diameter of wire $d_A=2.4m$
• Mass hanged with wire A $m_A=8kg$
• Young’s modulus of wire A Y₂ =1.37
• Length of wire B$l_B=2.8m$ l

• Diameter of wire B
• $$\begin{gathered} d_B=2×2.4mm \\ =4.8mm \end{gathered}$$
• Mass hanged with wire B $m=8kg$

Young’s modulus describes the relationship between stress (force per unit area) and strain (proportional deformation in an object). It does not depends on size or shape of an object.

As we know that young’s modulus is independent of size or shape. If the material is same then young’s modulus will be same. In question it give that material A and material B made of same material, so both will have same young’s modulus

$Y_B=Y_A$

Formula of young’s modulus is given as-

$$\begin{gathered} Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}} \\ =\frac{F.l}{a∆l} \\ =\frac{4mgl}{{\mathrm{Ï€»å}}^2∆\mathrm{l}} \end{gathered}$$

the amount of stretch in the wire B can be expressed as-

$$\begin{gathered} Y=\frac{4m_{B}g{l}_B}{{{\mathrm{Ï€»å}}_{\mathrm{B}}}^2∆{\mathrm{l}}_{\mathrm{B}}} \\ ∆l_B=\frac{4m_{B}g{l}_B}{{{\mathrm{Ï€»å}}_{\mathrm{B}}}^2{\mathrm{Y}}_{\mathrm{B}}} \end{gathered}$$

Substituting the values in the above equation,

$$\begin{gathered} ∆l_B=\frac{4×\left(8kg\right)×\left(9.8m/s^2\right)×\left(2.8\right)}{\mathrm{π}{\left(4.8×{10}^{-3}\mathrm{m}\right)}^2×1.37×{10}^{12}\mathrm{N}/{\mathrm{m}}^2} \\ =\frac{878.08kg{m}^2/s^2}{99.163×{10}^{6}N} \\ =8.85×{10}^{-6}kg{m}^2/s×\frac{1m}{1kgm/s^2} \\ =8.85×{10}^{-6}m \end{gathered}$$

Thus, the amount of stretch is $8.85×{10}^{-6}m$ and the Young’s modulus is $1.37×{10}^{12}N/m^2$.