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BITSAT 2011 Paper

Section: Physics

Question : 11 of 150

Marks: +1, -0

Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Validate

Solution:

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 =

\frac{\frac{F}{A}}{\frac{\Delta x}{l}}

... (i)

For wire 2 , Y =

\frac{\frac{F'}{3A}}{\frac{\Delta x}{\frac{l}{3}}}

... (ii)

From (i) and (ii) ,

\frac{F}{A} \times \frac{l}{\Delta x}

\frac{F'}{3A} \times \frac{l}{3\Delta x}

⇒ F' = 9F

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