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A brass wire is 2.0 m long at 40.02^@C a...

A brass wire is 2.0 m long at `40.02^@C` and is held taut with a little tension between the two rigid supports. The wire is now cooled down to `-47.5^@C`. If the tension developed in the wire is now T and the radius of the wire is 1.0 mm, find `(T)/(100) - 3 ` . Take, Coefficient of linear expansion of brass = `2.0 xx 10^(-5)""^@C^(-1)`
Young's modulus of brass =` 0.91xx 10^11 Pa`

Text Solution

Verified by Experts

The correct Answer is:
2
Here , `l = 2.0 , t_1 = 40.02^@C , t_2 = -47.50^@C`
`r = 1.0 mm = 10(-3) m`
`alpha = 2 xx 10^(-5) ""^@C^(-1) , Y = 0.91 xx 10^4 Pa`
` Delta l = alpha(t_2 - t_1) xx l`
Stress = strain x young.s modulus
` = alpha (t_2 - t_1) xx Y`
` = 2 xx 10^(-5) xx (-47.5 - 40.02 ) xx 0.91 xx 10^11`
` = 1.593 xx 10^8 Nm^(-2)`
Tension developed in the wire is
T = Stress x Area
`= 1.593 xx 10^8 xx 3.14 xx (1.0 xx 10^(-3))^2`
= 500 N
` (T)/(100) - 3 = 500/100 - 3= 5 - 3 = 2`
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