question_answer

A steel wire of 1.5 m long of diameter 0.25 cm and a brass wire of 1.0 m long of diameter 0.25 m are loaded as shown. Calculate, the elongation of the brass wire.

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                 \[\Delta {{L}_{Brass}}=\frac{{{(Mg)}_{Br}}\times {{L}_{Br}}}{(\pi {{r}^{2}})\times {{Y}_{Br}}}\]                 \[=\frac{58.8\times 1.0}{3.14\times {{(0.125\times {{10}^{-2}})}^{2}}\times 0.91\times {{10}^{11}}}\]                 \[=13\times {{10}^{-4}}m\]