AIEEE 2009 Solved Paper

As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section 3A will be ℓ∕3 (same volume as wire 1 ).
For wire 1 ,
y=‌

F∕A

∆x/ℓ

.....(i)
For wire 2.
‌Y=‌

F′/3A

∆x∕(ℓ∕3)

.........(ii)
From (i) and (ii),

F

A

×‌

ℓ

∆x

=‌

F′

3A

×‌

ℓ

3∆x

⇒F′=9F