Question

A steel wire is $1m$ long and $1m{m^2}$ in the area of cross-section. It takes $200N$ to stretch the wire by $1mm$, the force that will be required to stretch the wire of same material and cross-sectional area from a length of $10m$ to $1002cm$
(A) $100N$
(B) $200N$
(C) $400N$
(D) $2000N$

Answer 1

Hint
We need to first calculate the Young’s modulus of the wire in the first case by using the values given in the question. Then using that value of the Young’s modulus we can calculate the force that will be required in the second case.
In this solution, we will be using the following formula-
$\Rightarrow Y = \dfrac{{{F \mathord{\left/ {\vphantom {F A}} \right.} A}}}{{{{\Delta L} \mathord{\left/ {\vphantom {{\Delta L} L}} \right.} L}}}$
Where $Y$ is the Young’s modulus
$F$ is the force applied
$A$ is the area of cross-section of the wire
$L$ is the initial length of the wire and $\Delta L$ is the change in length due to the applied force.

Complete step by step answer
The Young’s modulus of a substance is its ability to withstand changes in length under lengthwise tension or compression. It is given by the formula,
$\Rightarrow Y = \dfrac{{{F \mathord{\left/ {\vphantom {F A}} \right.} A}}}{{{{\Delta L} \mathord{\left/ {\vphantom {{\Delta L} L}} \right.} L}}}$
In the question it is said that when a force of $F = 200N$ is applied on a wire of length $L = 1m$and cross-section $A = 1m{m^2} = {10^{ - 6}}{m^2}$, the change in the length of the wire is $\Delta L = 1mm = {10^{ - 3}}m$. So by substituting all these values in the formula for the Young’s modulus, we get its value as,
$\Rightarrow Y = \dfrac{{{{200} \mathord{\left/ {\vphantom {{200} {{{10}^{ - 6}}}}} \right.} {{{10}^{ - 6}}}}}}{{{{{{10}^{ - 3}}} \mathord{\left/ {\vphantom {{{{10}^{ - 3}}} 1}} \right.} 1}}}$
So we can write this in a simplified form as,
$\Rightarrow Y = \dfrac{{200}}{{{{10}^{ - 9}}}}$
This gives us the value of Young’s modulus of the wire as,
$\Rightarrow Y = 2 \times {10^{11}}N/{m^2}$
Now for the second case we are given the length of the wire as $L' = 10m$. The final length of the wire is given $1002cm = 10.02m$. Hence the increase in length is $\Delta L' = \left( {10.02 - 10} \right)m = 0.02m$. The cross-section of the wire will be the same, that is, $A = 1m{m^2} = {10^{ - 6}}{m^2}$. Now the young’s modulus will be same in both the cases, so we can write
$\Rightarrow Y = \dfrac{{{{F'} \mathord{\left/ {\vphantom {{F'} A}} \right.} A}}}{{{{\Delta L'} \mathord{\left/ {\vphantom {{\Delta L'} {L'}}} \right.} {L'}}}}$
Simplifying this formula we can write,
$\Rightarrow Y = \dfrac{{F'L'}}{{A\Delta L'}}$
Therefore we can write the formula for force by keeping only the $F'$ term in the LHS as
$\Rightarrow F' = \dfrac{{YA\Delta L'}}{{L'}}$
So by substituting all the values we can write,
$\Rightarrow F' = \dfrac{{2 \times {{10}^{11}} \times {{10}^{ - 6}} \times 0.02}}{{10}}$
On calculating the value in the numerator we have
$\Rightarrow F' = \dfrac{{4000}}{{10}}$
By doing the division, we get the value of the force as,
$\Rightarrow F' = 400N$
So the force required to stretch the wire will be $400N$.
Hence the correct answer is option (C).

Note
In both the cases the value of the Young’s modulus remains the same because the Young’s modulus is the characteristic property of the material of the wire. The young’s modulus can be calculated by the ratio of the longitudinal stress and the strain.