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 \[{{D}_{2}}=3\sqrt{\frac{7}{12}}\approx 2.3mm\].You need to login to perform this action.
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