A) \[l=10cm,D=1\text{ }mm\]
B) \[l=100\text{ }cm,\text{ }D=2mm\]
C) \[l=200\text{ }cm,\text{ }D=3\text{ }mm\]
D) \[l=300\text{ }cm,\text{ }D=4\text{ }mm\]
Correct Answer: B
Solution :
When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called Young's modulus of the material of the body. \[Y=\frac{stress}{strain}=\frac{F/A}{\Delta l/L}\] \[\Rightarrow \] \[\Delta l\propto \frac{1}{A}\propto \frac{1}{\pi {{r}^{2}}}\propto \frac{1}{{{D}^{2}}}\] \[\frac{1}{{{D}^{2}}}=\frac{10}{{{(1\times {{10}^{-1}})}^{2}}}=\frac{10}{{{10}^{-2}}}=1\times {{10}^{3}}\] \[\frac{1}{{{D}^{2}}}=\frac{100}{{{(2\times {{10}^{-1}})}^{2}}}=\frac{10}{4\times {{10}^{-2}}}=2.5\times {{10}^{3}}\] \[\frac{1}{{{D}^{2}}}=\frac{200}{{{(3\times {{10}^{-1}})}^{2}}}=22\times {{10}^{2}}=2.2\times {{10}^{3}}\] \[\frac{1}{{{D}^{2}}}=\frac{300}{{{(4\times {{10}^{-1}})}^{2}}}=18.75\times {{10}^{2}}=1.875\times {{10}^{3}}\] Hence, maximum extension is produced in second case, i.e., in option .
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