Young's modulus of rubber is \(10^4\text{ N/m}^2\) and area of cross-section is \(2\text{ cm}^2\). If force of \(2 \times 10^5\text{ dynes}\) is applied along its length, then length of wire becomes how much times of its initial length \(L\):- (Assume stress \(\propto\) strain)
Solution:
Converting Young's modulus to CGS units gives \(Y = 10^5\text{ dyne/cm}^2\). Stress is \(F/A = 2 \times 10^5 / 2 = 10^5\text{ dyne/cm}^2\). Since \(\text{Strain} = \text{Stress}/Y = 1\), we have \(\Delta L = L\). Thus, the final length becomes \(L + \Delta L = 2L\).
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