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Q. 11.9 A brass wire 1.8\; m long at 27^{\circ}C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39^{\circ}C,what is the tension developed in the wire if its diameter is 2.0\; mm\; ? ? Co-efficient of linear expansion\; of \; brass=2.0\times 10^{-5}K^{-1};  Young's\; modulus \; of\; brass=0.91\times 10^{11}Pa.

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Youngs Modulus of Brass, Y=0.91\times 10^{11}

Co-efficient of linear expansion of Brass, \alpha =2.0\times 10^{-5}K^{-1}

The diameter of the given brass wire, d = 2.0 mm

Length of the given brass wire, l = 1.8 m

Initial Temperature T1 = 27 oC

Final Temperature T2 = -39 o C

\\Y=\frac{Stress}{Strain}\\ Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\\ F=\frac{\Delta lAY}{l}\\ F=\frac{(l\alpha \Delta T)AY}{l}\\ F=\alpha \Delta TAY\\ F=\alpha \Delta TY\pi \frac{d^{2}}{4}\\ F=\frac{2.0\times 10^{-5}\times (-39-27)\times 0.91\times 10^{11}\times \pi \times (2\times 10^{-3})^{2}}{4}

F=-378N

The tension developed in the wire is 378 N. The negative sign signifies this tension is inward.

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