Question

On applying a stress of\[x{\text{ N/}}{{\text{m}}^2}\], the length of wire of some materials gets doubled. Value of Young’s modulus for the material of a wire is\[{\text{N/}}{{\text{m}}^2}\], is (assume Hooke's law to be valid and goes for approx. results).
A.$x$
B.$2x$
C.\[\dfrac{x}{2}\]
D.Insufficient information

Answer 1

Hint: In this question stress applied on a wire is given and when stress is applied then the length of the wire gets doubled so we first calculate the strain on the wire for the length change and since the stress is already is given so we can calculate the Young’s modulus of wire by using elasticity formula.

Step by step answer:
When stress of \[x{\text{ N/}}{{\text{m}}^2}\]is applied then the length of the wire gets doubled.
Stress applied=\[x{\text{ N/}}{{\text{m}}^2}\]
Young’s modulus of elasticity is given by the formula is given as \[\gamma = \dfrac{{Stress}}{{Strain}} - - (i)\]
Where strain is the ratio of change in length to the original length when stress is applied is given as
\[Strain = \dfrac{{\Delta l}}{l} - - (ii)\]
Now it is said that when stress of \[x{\text{ N/}}{{\text{m}}^2}\] is applied on a wire then the length of wire gets doubled, hence we can say the change in length is \[\Delta l = 2l - l\]
Original length\[ = l\]
Now we substitute these values in equation (ii) so we get,
\[Strain = \dfrac{{\Delta l}}{l} = \dfrac{{2l - l}}{l} = \dfrac{l}{l} = 1\]
Now we substitute the values of stress and strain is equation (i), so we get
Young’s modulus of elasticity
\[\gamma = \dfrac{{Stress}}{{Strain}} = \dfrac{x}{1} = x{\text{ N/}}{{\text{m}}^2}\]
Hence the value of Young’s modulus for the material of and wire is \[x{\text{ N/}}{{\text{m}}^2}\]

Option A is correct.

Note: Students must note that if the Young’s modulus of a material is large then the large amount of stress is required to create the same amount of strain.One must also keep in mind that if graph is given between the stress and strain the slope of that curve gives the value of Young's Modulus.