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A steel wire of diameter 0.5mm and Young modulus 2×1011Nm2 carries a load of mass M. The length of the wire with the load is 1.0m, A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which the main scale, of least count 1.0mm is attached. The 10 divisions of the Vernier scale correspond to 9 divisions of the main scale. Initially, the zero of the Vernier scale coincides with the zero of the main scale division. If the load on the steel wire is increased by 1.2kg, the vernier scale division which coincides with a main scale division is _________________.
[Take g=10ms2 and π=3.2 ]

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=FLAΔL

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
ΔL 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

d=5×104m

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

Y=2×1011Nm2

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=FLAΔL

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

ΔL=FLAY

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

A=πd24

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=10ms2

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

F=1.2×10=12N

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

ΔL=4FLπd2Y

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

ΔL=4×12×13.2×25×108×2×1011

On further solving, we get

ΔL=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=910MSD

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

L.C=1MSD1VSD
L.C=(1910)MSD=0.1mm

Hence, the 3rd 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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