91Ó°ÊÓ

The given data can be listed below as-

• The mass of the object hung from the copper wire is $10kg$.
• The wire stretches by an amount of .

The Young’s modulus is described as the product of the force applied and the length of an object divided by the product of the cross-sectional area and the change in the length of an object.

The equation of the Young’s modulus is helpful for solving the calculations.

The Young’s modulus of a wire made of a material is given by,

$Y=\frac{Fl}{A∆l}$

Where, Y is the Young’s modulus,$∆l$= change in length, F= force applied load (load), L = original length and A= Cross sectional area

The above equation can also be written as-

$∆l=\frac{Fl}{YA}$

So, from the above equation, it can be deduced that the change in the length is directly proportional to the load acted upon and the length of the wire and it is inversely proportional to the area of the wire.

In the first case, the stretch in the wire is expressed as-

$∆L∞\frac{1}{YA}$

…1)

a) If the mass is hung from the two-copper wire,

Then the force will become $F_2=\frac{F_1}{2}$

Here, the initial length is$8mm$.

For, $F_2=\frac{F_1}{F_2}$then the stretched length will become-

$$\begin{gathered} \frac{∆l_1}{∆l_2}=\frac{F_1}{F_2} \\ \frac{8mm}{∆l_2} \\ ∆l_2=4mm \end{gathered}$$

b)If the cross-sectional area of the two-copper wire is halved,

Then the area will become $A_2=\frac{A_1}{2}$

Here, the initial length isrole="math" localid="1653992127841" $8mm$.

For,$A_2=\frac{A_1}{2}$then the stretched length will become-

c)If the length of the two-copper wire is made twice,

Then the length will become

Here, the initial length is.

For,$A_2=\frac{A_1}{2}$then the stretched length will become-

Thus, a) the stretch will be less as the force is half and it will be less of about , b) If the cross-sectional area is made half aspiring the length same, stretching will be double which will be about and c) The stretching will be twice as the length is twice that will be about .