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A cylindrical steel wire of 3 m length ...

A cylindrical steel wire of `3 m` length is to stretch no more than `0.2 cm` when a tensile force of `400 N` is applied to each end of the wire. What minimum diameter is required for the wire ? `Y_(steel) = 2.1xx10^(11) N//m^2`

A

`2.91 mm`

B

`1.56 mm`

C

`1.91 mm`

D

`2.6mm`

Text Solution

Verified by Experts

The correct Answer is:
C
`Deltal = (Fl)/(AY) = (Fl)/((pid^(2)//4)Y)`
`:. d = sqrt((4 Fl)/(pi(Deltal)Y))`
` = sqrt((4xx400xx3)/(3.14xx0.2xx10^(-2)xx2.1xx10^(11)))`
`= 1.91 xx10^(-3) m = 1.91 mm`
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