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A decrease in temperature of 40^(@)C pro...

A decrease in temperature of `40^(@)C` produces a 0.1 % strain in a wire stretched between two fixed supports. If the area of cross section of the wire is `pi mm^(2)` and Young's modulus of its material is 160 GPa, the thermal stress in the wire is

A

16 Gpa

B

160 Mpa

C

`160 pi kPa`

D

`160 pi Pa`

Text Solution

Verified by Experts

The correct Answer is:
B
`(F)/(A)=Yxx" thermal strain"=(160xx10^(9))((0.1)/(100))=160xx10^(6) Pa`
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