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Young's modulus of rubber is 10^4 \mathrm{~N} / \mathrm{m}^2 and area of cross -section is 2 \mathrm{~cm}^2. If force of 2 \times 10^5 dynes is applied along its length, then its initial length l becomes-

Option: 1

3l


Option: 2

4l


Option: 3

2l


Option: 4

None of these


Answers (1)

best_answer

We have, 1 dyne =10^{-5} \mathrm{~N}
\Delta l=\frac{F \cdot l}{A Y}=\frac{2 \times 10^5 \times 10^{-5} \times l}{2 \times 10^{-4} \times 10^4}=l

Thus new length =l+\Delta l=l+l \Rightarrow 2 l 

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