BITSAT 2011 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 crosssection 3A will be l/3 (same volume as wire 1).
For wire 1, Y =

F∕A

Δx∕l

... (i)
For wire 2 , Y =

F′∕3A

Δx∕(l∕3)

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

F

A

×

l

Δx

=

F′

3A

×

l

3Δx

⇒ F' = 9F