Question

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm are connected end to end. When stretched by a force, the elongation in length 0.50 mm is produced in the copper wire. The stretching force is
(Ycu​=1.1×10$^{11}$N/m2,Ysteel​=2.0×10$^{11}$ N/m2)
A. 5.4$\times $10$^{2}$ N
B. 3.6$\times $10$^{2}$ N
C. 2.4$\times $10$^{2}$ N
D. 1.8$\times $10$^{2}$ N

Answer 1

Hint: At first find the radius from the diameter given. Then write the equation of young’s modulus for both the wires. Now, consider a force in both the wires. Then write the equation for young’s modulus with respect to force.

Complete answer:
We know that the length of the copper wire is,
l$_{1}$ ​=2.2m
And we also, know that the length of the steel wire is,
 l$_{2}$ =1.6m
The elongation that occur in length is,

△l=0.5 mm
    =0.5×10$^{3}$m
 Radius of the Copper wire is,
 (radius = Diameter/2)
${{r}_{1}}$ =1.5×10$^{3}$m
Young’s modulus for copper wire,
Y$_{1}$​=1.1×10\[^{11}\]N/m$^{2}$
Young’s modulus for steel wire,
Y$_{2}$ ​=2.0×10\[^{11}\]N/m$^{2}$

Let the stretching force in both the wires be F then,
for Copper wire,

Y$_{1}$​=$\dfrac{F}{\pi r_{1}^{2}}\times \dfrac{{{l}_{1}}}{\Delta {{l}_{1}}}$ ​​

⇒F​=$\dfrac{{{Y}_{1}}\pi r_{1}^{2}\times \Delta {{l}_{1}}}{{{l}_{1}}}$ ​​
=$\dfrac{1.1\times {{10}^{-11}}}{2.2}\times \dfrac{22}{7}\times {{(1.5\times {{10}^{-3}})}^{2}}\times 0.5\times {{10}^{3}}$
=1.8×10$^{2}$N
So the stretching force in the copper wire be 1.8×10$^{2}$N.

So, the correct answer is “Option D”.

Additional Information:
The ability of a material to withstand changes in length when it is stretched, is known as the modulus of elasticity,
Young's modulus is the ratio of longitudinal stress and strain.

Note:
Convert all the given units to a single unit. Write the equation for young’s modulus properly in proper order, Young's modulus is the ratio of longitudinal stress and strain. Compare the equation of ${{Y}_{1}}$ with F.