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Question: The radii and Young’s moduli of two uniform wires A and B are in the ratio 2 : 1 and 1 : 2 respectiv...

The radii and Young’s moduli of two uniform wires A and B are in the ratio 2 : 1 and 1 : 2 respectively. Both wires are subjected to the same longitudinal force. If the increase in length of the wire A is one percent, the percentage increase in length of the wire B is

A

1.0

B

1.5

C

2.0

D

3.0

Answer

2.0

Explanation

Solution

: Young’s modulus, Y=FLπr2ΔLY = \frac{FL}{\pi r^{2}\Delta L}

or ΔLL=Fπr2Y\frac{\Delta L}{L} = \frac{F}{\pi r^{2}Y}

for the same force.

ΔLBLBΔLALA=(rArB)2(YAYB)=(21)2(12)=2\therefore\frac{\frac{\Delta L_{B}}{L_{B}}}{\frac{\Delta L_{A}}{L_{A}}} = \left( \frac{r_{A}}{r_{B}} \right)^{2}\left( \frac{Y_{A}}{Y_{B}} \right) = \left( \frac{2}{1} \right)^{2}\left( \frac{1}{2} \right) = 2

ΔLBLB=2(ΔLALA)\frac{\Delta L_{B}}{L_{B}} = 2\left( \frac{\Delta L_{A}}{L_{A}} \right)

ΔLBLB×100=2(ΔLALA×100)=2%\frac{\Delta L_{B}}{L_{B}} \times 100 = 2\left( \frac{\Delta L_{A}}{L_{A}} \times 100 \right) = 2\%