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 theradius 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

A brass wire is 2.0 m long at 40.02^@C and is held taut with a little

1.	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 theradius 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
Answer» Solution :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`