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A 4.0 m long copper wire of cross sectio...

A 4.0 m long copper wire of cross sectional area `1.2 cm^(2)` is stretched by a force of `4.8 × 10^(3)` N stress will be -

A

`4.0xx10^(7)" N/mm"^(2)`

B

`4.0xx10^(7)" KN/m"^(2)`

C

`4.0xx10^(7)" N/m"^(2)`

D

None

Text Solution

Verified by Experts

The correct Answer is:
C
Stress `=(F)/(A) =(4.8xx10^(3)N)/(1.2xx10^(-4)m^(2))=4.0xx10^(7)" N/m"^(2)`
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