A steel wire of `2.0 mm^2` cross- section is held straight(but under no tension) by attaching it firmly to two points a distance `1.50 m` apart at `30^@ C`. If the temperature now decreases to `-10^@ C` and if the two points remain fixed, what will be the tension in the wire ? For steel, `Y = 20.0000 MPa`.

Answer

A, B

Explanation

Strain = `(Delta l)/(l) = (alpha Delta theta)`
Stress = `Y xx "strain" = gamma alpha Delta theta`
`:.` Tension = `(A)` (stress) `= Y A alpha Delta theta`
Substituting thr value, we get
`T = Y A alpha Delta theta`
= `(2.0 xx 10^11)(2 xx 10^-6)(1.2 xx 10^-5)(40)`
= `192 N`.