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Young's modulus of rubber is 10^(4) N//m...

Young's modulus of rubber is `10^(4) N//m^(2)` and area of cross section is `2 cm^(2)`. If force of `2 xx 10^(5) dyn` is applied along its length, then its final length becomes

A

3L

B

4L

C

2L

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
C
`Y= 10^(4) N//m^(2) , A = 2 xx 10^(-4) m^(2), F = (20 xx 10^(5))/(10^(5)) = 2N`
`Delta l= (Fl)/(YA) = (2 xx L)/(10^(4) xx 10^(-4) xx 2) = L`
Final length `= L + L= 2L`
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