Answer

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**Hint:**In order to solve this question you have to remember all the concepts related to Young’s modulus. Basically, a modulus is a numerical value that represents a physical property of a material and Young’s modulus is used to represent how easy it is to deform a material. Also, you have to remember the formula for finding the young’s modulus.

**Formula used:**

The formula for Young’s modulus is given by,

$Y = \dfrac{{FL}}{{A\Delta L}}$

Where $Y$ is the Young’s modulus

$F$ is the force applied

$L$ is the length of the material before applying force

$A$ is the area of cross-section of the material wire

$\Delta L$ is the change in the length of the material after applying the force

**Complete step by step solution:**

Here, in the question, the diameter of the wire is given by,

$d = 0.5mm$

On converting it into its standard units that are in meters, we have

$ \Rightarrow d = 5 \times {10^{ - 4}}m$

The value of the Young’s modulus of the wire is also given in the question by,

$Y = 2 \times {10^{11}}N{m^{ - 2}}$

And also the length of the wire before applying force is given in the question by,

$L = 1.0m$

We have all the values, now apply the formula for the Young’s modulus, we get

$Y = \dfrac{{FL}}{{A\Delta L}}$

Now from the above relation find the value of change in length of wire after the force is applied

$\Delta L = \dfrac{{FL}}{{AY}}$

As we know that the area of the wire is given by,

$A = \dfrac{{\pi {d^2}}}{4}$

And the force applied is given by,

$F = mg$

Here the mass of the load on the steel wire is given by,

$m = 1.2kg$

And also the value of acceleration due to gravity is given by,

$g = 10m{s^{ - 2}}$

On putting the value mass and acceleration due to gravity, we get the value of force applied

$F = 1.2 \times 10 = 12N$

Now put the expression for the area of the wire in the above relation of the Young’s modulus, we get

$ \Rightarrow \Delta L = \dfrac{{4FL}}{{\pi {d^2}Y}}$

Now on putting all the available values in the above expression, we get

$ \Rightarrow \Delta L = \dfrac{{4 \times 12 \times 1}}{{3.2 \times 25 \times {{10}^{ - 8}} \times 2 \times {{10}^{11}}}}$

On further solving, we get

$ \Rightarrow \Delta L = 0.3mm$

As in the question, it is also given that the \[10\] divisions of the Vernier scale are equal to \[9\] divisions of the main scale

$10VSD = 9MSD$

So the one division of the vernier scale is equal to,

$1VSD = \dfrac{9}{{10}}MSD$

We also know that the least count of vernier caliper is given by

$L.C = 1MSD - 1VSD$

$ \Rightarrow L.C = \left( {1 - \dfrac{9}{{10}}} \right)MSD = 0.1mm$

**Hence, the ${3^{rd}}$ vernier scale division coincides with the main scale division.**

**Note:**The Young’s modulus compares the tensile strain with the tensile stress. Young’s modulus is also known as the modulus of elasticity in tension, it is a mechanical property which determines the tensile stiffness of the material. Also don’t confuse material stiffness with material strength or hardness or toughness, each of them is different in terms of properties of the material.

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