Question

Two wires are made of the same kind of metal. Wire A has a diameter of 2.4 mm and is initially 2.8m long. You have a 8kg mass from wire A, measure the amount of stretch, and determine Young’s modulus to be Wire B, which is made of the same kind of metal as wire A, has the same length as wire A but twice the diameter. You hang the same 8kg mass from wire B, measure the amount of stretch, and determine Young’s modulus $Y_B$,

Which one of the following is true?

Short answer

1..$Y_B=Y_A$

The amount of stretch isrole="math" localid="1653995796689" $8.85\times {10}^{-6}m$ and the Young’s modulus is $1.37\times {10}^{12}N/m^2$.

Step-by-step solution

Identification of the given data

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

Significance of the Young’s modulus of a wire

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.

Determination of Young’s Modulus of wire B

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

$Y_B=Y_A$

Determination of amount of stretch of wire B

Formula of young’s modulus is given as-

$$\begin{gathered} Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}} \\ =\frac{F.l}{a∆l} \\ =\frac{4mgl}{{\mathrm{\pi d}}^2∆\mathrm{l}} \end{gathered}$$

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

$$\begin{gathered} Y=\frac{4m_{B}g{l}_B}{{{\mathrm{\pi d}}_{\mathrm{B}}}^2∆{\mathrm{l}}_{\mathrm{B}}} \\ ∆l_B=\frac{4m_{B}g{l}_B}{{{\mathrm{\pi d}}_{\mathrm{B}}}^2{\mathrm{Y}}_{\mathrm{B}}} \end{gathered}$$

Substituting the values in the above equation,

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

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