Question

A steel wire 1 m long and $1m{m^2}$ in the area of cross-section. If it takes 200N to stretch this wire by 1 mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal of 10 m to a length of 1002 cm:
A) 1000 N.
B) 200 N.
C) 400 N.
D) 2000 N.

Answer 1

The stress is defined as the ratio of force and area of the cross-section. The strain is defined as the ratio of change of length and the original length. Young’s modulus is defined as the ratio of stress and strain.

Formula used:
The formula of the Young’s modulus is given by,
$\gamma = \dfrac{\sigma }{\varepsilon }$
Where, Young’s modulus is given by $Y$ the stress is given by $\sigma$, and strain is given by$\varepsilon$.

Complete step by step answer:
It is given in the problem of a steel wire 1 m long and$1m{m^2}$ in the area of cross-section. If it takes 200N to stretch this wire by 1 mm and we need to find the force that will be required to stretch a wire of the same material as well as diameter from its normal of 10 m to a length of 1002 cm.
First of all let us calculate the value of the Young’s modulus of the material.
The formula of the Young’s modulus is given by,
$\gamma = \dfrac{\sigma }{\varepsilon }$
Where the Young’s modulus is given by $\gamma$ the stress is given by $\sigma$ and strain is given by $\varepsilon$.
The force applied is equal to 200 N the area of cross-section is $1m{m^2}$, length of wire is 1 m and the change in length is equal to 1 mm. Therefore the Young’s modulus is given by,
$\Rightarrow \gamma = \dfrac{\sigma }{\varepsilon }$
$\Rightarrow \gamma = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$
$\Rightarrow \gamma = \dfrac{{\left( {\dfrac{{200}}{{{{10}^{ - 6}}}}} \right)}}{{\left( {\dfrac{{{{10}^{ - 3}}}}{1}} \right)}}$
$\Rightarrow \gamma = 2 \times {10^{11}}$
The force required to stretch a wire of the same material with area same but length equal to 10 m and is stretched to 1002 cm is equal to.
$\Rightarrow F = \dfrac{{\gamma A\left( {\Delta l} \right)}}{l}$
$\Rightarrow F = \dfrac{{\left( {2 \times {{10}^{11}}} \right)\left( {{{10}^{ - 6}}} \right)\left( {1002 - 1000} \right) \times {{10}^{ - 2}}}}{{10}}$
$\Rightarrow F = 400N$.

The force that stretches the same material from 10 m to 1002 cm is given by $F = 400N$. The correct answer is option C.

Note:
Young’s modulus is constant for a material and is different for different materials but the stress and strain that acts upon the different material will be different depending upon Young’s modulus of that material.