question_answer

A 5m aluminum wire  \[(y=7\times {{10}^{10}}N/{{m}^{2}})\]of diameter 3mm supports a 40kg mass. In order to have the same elongation in a copper wire \[(y=12\times {{10}^{10}}N/{{m}^{2}})\] of the same length under the same wieght, the diameter should be in mm:        

A)  \[1.75\]              

B)  \[4.0\]                      

C)  \[2.3\]               

D)  \[5.0\]                      

Correct Answer: C

Solution :

If strain is small, the ratio of the longitudinal to the corresponding longitudinal strain is called Young's modulus (Y) of the material of the body. \[Y=\frac{Stress}{Strain}=\frac{F/A}{l/L}=\frac{F.L}{\pi {{r}^{2}}l}\] Here, \[{{Y}_{1}}=7\times {{10}^{10}}N/{{m}^{2}}.\] \[{{Y}_{2}}=12\times {{10}^{10}}N/{{m}^{2}}.\] \[{{r}_{1}}=\frac{{{D}_{1}}}{2}=\frac{3}{2}mm,\,\,{{r}_{2}}=\frac{{{D}_{2}}}{2}\] \[\therefore \] Taking ratio of Young's modulus and putting \[r=\frac{D}{2},\] we get \[\frac{{{Y}_{2}}}{{{Y}_{1}}}={{\left( \frac{{{D}_{1}}}{{{D}_{2}}} \right)}^{2}}\] \[\Rightarrow \] \[\frac{12\times {{10}^{10}}}{7\times {{10}^{10}}}={{\left( \frac{3}{{{D}_{2}}} \right)}^{2}}\] or \[\frac{3}{{{D}_{2}}}=\sqrt{\frac{12}{7}}\] or \[{{D}_{2}}=3\sqrt{\frac{7}{12}}\approx 2.3mm\].