Young's modulus of rubber is ${10^4}\,N/{m^2}$ and area of cross-section is $2\,c{m^2}$. If force of $2 \times {10^5}$ dynes is applied along its length, then its initial length $l$ becomes
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(c) $Y = {10^4}N/{m^2},A = 2 \times {10^{ - 4}}{m^2},F = 2 \times {10^5}dyne = 2N$
$l = \frac{{FL}}{{AY}} = \frac{{2 \times L}}{{2 \times {{10}^{ - 4}} \times {{10}^4}}} = L$
Final length $=$ initial length $+$ increment $= 2L$
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