question_answer 10)

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2 m long. These at each end are of copper and middle one is of iron. Determine the ratio of their diameters if each is to have same tension. Youngs modulus of elasticity for copper and steel are \[\frac{1}{200}\] \[\frac{\text{force}}{\text{area}}\text{=}\frac{\text{mg}}{\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}\text{=}\frac{\text{50 }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 8}}{\left( \text{22/7} \right)\text{ }\!\!\times\!\!\text{ }{{\left( \text{1/200} \right)}^{\text{2}}}}\] and  \[=6\cdot 24\times {{10}^{6}}N{{m}^{-2}}.\]\[{{m}^{-3}}\] respectively.

Answer:

As each wire has same tension F, so each wire has same extension due to mass of rigid bar. As each wire of same length, hence each wire has same strain. If D is the diameter of wire, then                  \[h=\frac{0\cdot 76\times 13\cdot 6\times {{10}^{3}}\times 9\cdot 8}{984\times 9\cdot 8}=10\cdot 5m.\] or \[{{10}^{9}}\] \[={{10}^{9}}\] \[\text{3 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{2}}}\]\[3\cdot 5\]                 \[\text{4 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{2}}}\]