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A steel wire of uniform cross-section `1 mm^(2)` is heated to `70^(@)C` and stretched by tying it two ends rigidly. Calculate the change in tension on the wire when temperature falls form `70^(@)C` to `35^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
77
Here A=1 `m m^(2)= 10^(-6) m^(2)`
`Delta T= 70- 35= 35^(@)`
`alpha= 1.1 xx 10^(-5) ""^(@)C^(-1), Y= 2.0 xx 10^(11) Nm^(-2)`
increase in length `Delta l= l alpha Delta T`
`therefore` Strain `= (Delta l)/(l)= alpha Delta T= 1.1 xx 10^(-5) xx 35= 38.5 xx 10^(-5)`
If T is the tension in the wire due to the decrease in temperature, then
Stress `= (T)/(A)= (T)/(10^(-6))Nm^(-2)`
But Stress= Y `xx` Strain
`therefore (T)/(10^(-6))= 2.0 xx 10^(11) xx 38.5 xx 10^(-5)`
`or T= 2.0 xx 38.5 = 77.0 N`
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