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If the ratio of diameters, lengths and young’s modulus of steel and copper wires shown in the figure are p,q and s respectively, then the corresponding ratio of the increase in their lengths would be ________
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Answer
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Hint
By using the formula of the Young’s modulus of the wire we can find the formula for the increase in the length. Then we can take the ratio of the increase in length for both the wires. We can substitute the given values of the ratio of diameter, lengths and Young’s modulus. And the force on each wire will be the amount of weight on each of the wires.

Formula Used: In this solution we will be using the following formula,
 Y=FLAΔL
where Y is the young’s modulus
 F is the force
 L is the length
 A is the area and
 ΔL is the change in the length.

Complete step by step answer
The young’s modulus of a wire is given by the formula,
 Y=FLAΔL
Here A is the area of the cross section of the wire. So we can write the area of cross section in the term of diameter of the wire as,
 A=π(D2)2
Hence we get,
 A=πD24
So substituting this value in the formula we get,
 Y=4FLπD2ΔL
Now we can bring the ΔL to the LHS and take the Y to the RHS. Therefore we get,
 ΔL=4FLπD2Y
Now let us consider the force, length, diameter and the young’s modulus of the steel wire be, Fs, Ls, Ds and Ys respectively. Therefore, the change in length of the steel wire is,
 ΔLs=4FsLsπDs2Ys
And for the copper wire let the force, length, diameter and the young’s modulus be, Fc, Lc, Dc and Yc respectively. Therefore the change in length of the copper wire is,
 ΔLc=4FcLcπDc2Yc
Now we can take the ratio of the change in lengths of the steel and copper wire as,
 ΔLsΔLc=4FsLsπDs2Ys4FcLcπDc2Yc
So the 4 and the π gets cancelled and we can simplify the form of this equation as,
 ΔLsΔLc=FsFc×LsLc×Dc2Ds2×YcYs
Now according to the diagram, the force on the steel wire is the sum of the weights of the two blocks. So we get, Fs=(2m+5m)g=7mg
And the force on the copper wire is due to the weight of the second block only. Hence we get,
 Fc=5mg
In the question we are given the ratio of the lengths as, LsLc=q
The ratio of the diameters is given as, DsDc=p and the ratio of the young’s modulus is given as, YsYc=s. Now substituting all the values in the equation we get,
 ΔLsΔLc=7mg5mg×q×(1p)2×1s
Hence on simplifying and cancelling the like terms we get,
 ΔLsΔLc=7q5p2s
This is the corresponding ratio of the increase in lengths of the steel wire to the copper wire.

Note
The young’s modulus of a wire is the property of the material of a wire. This is the mechanical property that measures the tensile stiffness of a solid material. It can be given by the ratio of the tensile stress and the axial strain.
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