Two wires A and B are of the same length. The diameter is in the ratio 1: 2 and the youngs modulus are in the ratio of 2: 1. If they are pulled by the same force then their elongation will be in the ratio

Answers (1)

using Y=\frac{stress}{strain}

we get

Y=\frac{FL}{A\Delta l}
 

where

A = Area of wire

L = Original length

\Delta l  = Change in length

F = Force

now after rewriting it we get  

 \frac{\Delta l}{l}= \frac{F}{AY}\: \: or \: \: \Delta l= \frac{Fl }{\pi r^2Y}\\\\\\ \frac{\Delta l_1}{\Delta l_2} = \frac{Y_2r_{2}^{2}}{Y_1r_{1}^{2}}= \frac{1}{2}\times \frac{4}{1}= 2: 1

 

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