A mass of 2kg is suspendedfrom a fixed point by a wire of length 3m and diameter 0.5 mm. Initially the wire is just unstretched, the mass resting on a fixed support. By hoH' much must the temperature fall if the mass is to be entirely supported by the wire? Given that Y of the wire = 206 G Pa, alpha = 11 xx 10^(-6)//^(0) C .

A mass of 2kg is suspendedfrom a fixed point by a wire of length 3m an

1.	A mass of 2kg is suspendedfrom a fixed point by a wire of length 3m and diameter 0.5 mm. Initially the wire is just unstretched, the mass resting on a fixed support. By hoH' much must the temperature fall if the mass is to be entirely supported by the wire? Given that Y of the wire = 206 G Pa, alpha = 11 xx 10^(-6)//^(0) C .
Answer» Solution :When the temperature DECREASES the wire contracts in length and LIFTS the mass up. But the weight begins to produce a stress and counteracts the contraction PRODUCED. Hence the contraction due to cooling is equal to the stretching produced by the weight W. `therefore DELTA L = (WL)/(AY) = (2 xx 9.8 xx 3)/(pi (0.25)^(2) xx 10^(-6) xx 206 xx 10^(9))` Now the contraction due to cooling ` = L alpha d theta = 3 xx 11 xx 10^(-6) xx d theta` Since the contraction due to cooling = extension produced by the weight. On solving `d theta = 44^(@)`C