Solveeit Logo
Login

Question

Question: A steel wire of length \(1m\) has cross-sectional area \(1c{m^2}\).If Young’s modulus of steel is \(...

A steel wire of length 1m1m has cross-sectional area 1cm21c{m^2}.If Young’s modulus of steel is 1011Nm2{10^{11}}N{m^2}, then the force required to increase the length of wire by 1mm1mm will be:
A. 1011N{10^{11}}N
B. 107N{10^7}N
C. 104N{10^4}N
D. 102N{10^2}N

Explanation

Solution

When the length of an object is changed to some extent, Young’s modulus is calculated. It is the measure of elasticity of that object. To change the length, force is required.

Formula used:
E=(FA)(lL)E = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{l}{L}} \right)}}
⇒E=FLAl\Rightarrow E = \dfrac{{FL}}{{Al}}
From this, we can understand that the ratio of longitudinal stress and longitudinal strain is known as the young’s modulus.
Where,
EE=Young’s modulus,
FF=force,
AA=cross sectional area,
ll=change in length,
LL=length

Complete step by step answer:
E=1011Nm2E = {10^{11}}\dfrac{N}{{{m^2}}}
L=1mL = 1m
A=1cm2A = 1c{m^2}
⇒(1×10−2)2m2\Rightarrow {\left( {1 \times {{10}^{ - 2}}} \right)^2}{m^2}
⇒1×10−4m2\Rightarrow 1 \times {10^{ - 4}}{m^2}
l=1mml = 1mm
⇒1×10−3m\Rightarrow 1 \times {10^{ - 3}}m
We made changes in the given data, as the solution can be found easily while all the data is in the same unit. That’s why we change cm&mmcm\& mm into mm.
The formula we know is,
E=FLAlE = \dfrac{{FL}}{{Al}}
The required value is force, we convert this equation by keeping FF on one side and reciprocating other terms into the opposite side.
∴F=EAlL\therefore F = \dfrac{{EAl}}{L}
Now the equation is ready to apply the values,
F=1011×10−4×10−3F = {10^{11}} \times {10^{ - 4}} \times {10^{ - 3}}
⇒F=1011−(3+4)\Rightarrow F = {10^{11 - \left( {3 + 4} \right)}}
⇒1011−7\Rightarrow {10^{11 - 7}}
⇒104\Rightarrow {10^4}
∴F=104N\therefore F = {10^4}N
(NN,newton is the unit of force)

Therefore, the correct option is option C.

Additional information:
(i)To find elasticity across the length, we find young modulus. It is the ratio of longitudinal stress to longitudinal strain.
(ii)Longitudinal stress is the ratio of force to area. Whereas in longitudinal strain, it is the ratio of change in length to the original length.
(iii) We not only have young modulus. We have rigidity modulus and the bulk modulus. Rigidity modulus is the measure of elasticity in the case of the area changed. Bulk modulus is the measure of elasticity in case the volume changed.

Note:
The dimensional formula for modulus of elasticity is ML−1T−2M{L^{ - 1}}{T^{ - 2}}. Its unit is Nm2\dfrac{N}{{{m^2}}}. There are three types of modulus. They are the Modulus of elasticity or Young’s modulus, Bulk modulus, and Rigidity modulus.