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Question: A wire of length 1 m and radius 1 mm is subjected to a load. The extension is 'x'. The wire is melte...

A wire of length 1 m and radius 1 mm is subjected to a load. The extension is 'x'. The wire is melted and then drawn into a wire of 1 mm square cross-section. What is its extension under the same load?
A) π2x{\pi ^2}x
B) πx2\pi {x^2}
C) πx\pi x
D) πx\dfrac{\pi }{x}

Explanation

Solution

The shape of wire is changed from cylindrical to square hence we can easily find out the length of square wire with the help of equating volume of both as equal. Then force per unit area is directly proportional to change in length per unit length and both have the same Young’s modulus hence again equating together we will find the change in length.

Step by step solution:
Step 1:
The question is based on the change in length with respect to the previous one and this is stated in Young’s modulus. Let’s see a brief definition.
Young's modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. Sometimes referred to as the modulus of elasticity, Young's modulus is equal to the longitudinal stress divided by the strain.
Step 2:
Let the initial length of the wire be L=1m and radius r=1mm, converting it into meter then r=10−3m{10^{ - 3}}m
A load in put on the wire and because of which there is an extension be ΔL2\Delta {L_2} =x
Wire after melting and casted into square cross-section of side 1mm or 1×10−3m1 \times {10^{ - 3}}m
Area of cross- section will be A=1×10−6m1 \times {10^{ - 6}}m
Now in both conditions the volume should be the same.
Before melting it was in a shape cylinder and after melting it was in the shape of a square.
Initial volume= final volume ⇒\Rightarrow πr2L\pi {r^2}L =A× (L2{L_2})….. (1)
We have the value of radius and length L and A and substitute in (1) we get L2{L_2} (change in length)
This implies L2{L_2}=πr2LA\dfrac{{\pi {r^2}L}}{A} ⇒π×10−6×110−6 \Rightarrow \dfrac{{\pi \times {{10}^{ - 6}} \times 1}}{{{{10}^{ - 6}}}}
Which meansL2{L_2}equals to π\pi
Initial Young’s modulus should be equal to final Young’s modulus of the wire then, Y1=Y2{Y_1} = {Y_2}
This implies FLAx\dfrac{{FL}}{{Ax}} =FL1A1ΔL2\dfrac{{F{L_1}}}{{{A_1}\Delta {L_2}}} and F will get cancel out.1πr2×(x)\dfrac{1}{{\pi {r^2} \times \left( x \right)}} =π10−6\dfrac{\pi }{{{{10}^{ - 6}}}}
From here we will get ΔL\Delta L (change in length equal to π2x{\pi ^2}x

So option A is correct.

Note: All the things are explained in the hint and solution part. Important thing is the Young’s formula and the reason behind it once it is read then any question based on the change in length with respect to the original one or the force per unit area will be easy to solve.