The force required to stretch a wire of cross-section $1 \mathrm{~cm}^{2}$ to double its length will be : (Given Yong's modulus of the wire $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$)
Solution
<p>$A = 1$ cm<sup>2</sup></p>
<p>$Y = {{Fl} \over {A\Delta l}}$</p>
<p>$$F = {{YA\Delta l} \over l} = {{2 \times {{10}^{11}} \times {{10}^{ - 4}} \times l} \over l}$$</p>
<p>$= 2 \times {10^7}$ N</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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