Explanation
$$F\propto \Delta l$$
$$F_1\propto l_1-l$$
Where $$l$$, is the original length of the wire
$$F_2\propto l_2-l$$
$$\dfrac{F_1}{F_2}=\dfrac{l_1-l}{l_2-l}$$
$$F_1l_2-F_1l=F_2l_1-F_2l$$
$$(F_2-F_1)l=F_2l_1-F_1l_2$$
$$l=\dfrac{F_2l_1-F_1l_2}{F_2-F_1}$$
Hint: Young's Modulus is the ratio of stress and strain.
Solution:
Step1: Observe the given diagram,
Step:2 Use Young’s modulus equation
Using Young’s modulus equation we have,
$$Y = \dfrac{{F \times l}}{{A \times \Delta l}}.....(1)$$
Where, Y=Young’s modulus, F=Applied force, l=Initial length, A=Square area & ΔL= Elongation.
So, we have slope, $$\dfrac{{F(load)}}{{\Delta l(elongation)}} = \dfrac{{Y \times A}}{l}$$
Hence, $$slope \propto A$$ or $$slope \propto {r^2}$$
It is clear that, for thinnest wire slope is minimum i.e. wire ‘OA’ is thinnest.
Hence, option (A) is correct.
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