A) 1.75
B) 2.0
C) 2.3
D) 5.0
Correct Answer: C
Solution :
When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called Youngs modulus \[(Y)\]of the material of the body. \[Y=\frac{stress}{strain}=\frac{F/A}{l/L}=\frac{F\cdot L}{\pi {{r}^{2}}l}\] Given, \[{{Y}_{1}}=7\times {{10}^{10}}N/{{m}^{2}}\] \[{{Y}_{2}}=12\times {{10}^{10}}N/{{m}^{2}}\] \[{{r}_{1}}=\frac{{{D}_{1}}}{2}=\frac{3}{2}mm\], \[{{r}_{2}}=\frac{{{D}_{2}}}{2}\] \[\therefore \] \[\frac{{{Y}_{2}}}{{{Y}_{1}}}={{\left( \frac{{{D}_{1}}}{{{D}_{2}}} \right)}^{2}}\] \[\frac{12\times {{10}^{10}}}{7\times {{10}^{10}}}={{\left( \frac{3}{{{D}_{2}}} \right)}^{2}}\] \[\Rightarrow \] \[\frac{3}{{{D}_{2}}}=\sqrt{\frac{12}{7}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{D}_{2}}=3\sqrt{\frac{7}{12}}\,\approx \,2.3\,mm\]You need to login to perform this action.
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