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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 these

Text Solution

Verified by Experts

The correct Answer is:
C
(c) Elongation of the wire
`Deltal=(Fl)/(AY)=((2)l)/((10^(4))(2xx10^(-4)))=l" "[because 1 "dyne"=10^(-5) N]`
`therefore` New length will become 2l.
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