A) \[1.1\times {{10}^{-4}}m\]
B) \[1.3\times {{10}^{-4}}m\]
C) \[1.5\times {{10}^{-4}}m\]
D) \[1.7\times {{10}^{-4}}m\]
Correct Answer: B
Solution :
If a wire of length L and radius r increase in length by \[\Delta L\] under a suspended load Mg, then the Youngs modulus of the material of the wire is \[Y=\frac{Mg\,L}{(\pi {{r}^{2}})\times \Delta L}\] or \[\Delta L=\frac{Mg\,L}{(\pi {{r}^{2}})Y}\] For the brass wire, the suspended load is \[6.0\,kg\times 9.8\,N/kg\] \[=58.8\,N\] \[\therefore \] Elongation of the brass wire is given by \[\Delta {{L}_{Brass}}=\frac{{{(Mg)}_{Br}}\times {{L}_{Br}}}{(\pi {{r}^{2}})\times {{Y}_{Br}}}\] \[=\frac{58.8\times 1.0}{3.14\times {{(0.125\times {{10}^{-2}})}^{2}}\times 0.91\times {{10}^{11}}}\] \[=13\times {{10}^{-4}}m\]You need to login to perform this action.
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