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Fig. shows a wire of length l and of slightly non-uniform cross-section. Its radius at one end is R₁ and at the other, it is R₂
The wire is stretched by forces F and F applied along the length in opposite directions and normal to the ends. Y being the Young's rpodulus, extension caused in the wire is nearly:

.

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$\frac{FlY}{{πR}_{1}^{2} R_{2}^{2}}$

$\frac{FlY}{R_{1} R_{2}}$

$\frac{Fl}{{πYR}_{1} R_{2}}$

$\frac{FY}{{lπR}_{1} R_{2}}$

answer is A.

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Detailed Solution

Effective area of the wire = $π R_{1} R_{2}$

so extension in the wire =e= $\frac{F l}{Y π R_{1} R_{2}}$

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