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A wire of length 1 m and its area of cr...

A wire of length 1 m and its area of cross-section is 1 `cm^(2)` and Young's modulus of the wire is `10^(11)N//m^(2)`. Two forces each equal to F are applied on its two ends in the opposite directions. If the changein length is 1 mm, then the value of F will be

A

`0.5xx10^(4)N`

B

`10^(4)N`

C

`2xx10^(4)N`

D

`10^(8)N`

Text Solution

Verified by Experts

The correct Answer is:
B
`y=(FL)/(Al)`
`therefore F=(yAl)/(L)=(10xx10^(11)xx10^(-4)xx10^(-3))/(1)=10^(4)N`
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