| \[{{Y}_{s}}\,(steel)=2.0\times {{10}^{11}}\,N/{{m}^{2}}\] |
| \[{{Y}_{c}}\,(copper)=1.2\times {{10}^{11}}\,N/{{m}^{2}}\] |
A) \[{{l}_{s}}=\,\,0.75\text{ }cm,\,\,\,{{l}_{c}}\,=\,\,1.25\text{ }cm\]
B) \[{{l}_{s}}=\,\,1.25\text{ }cm,\,\,\,{{l}_{c}}\,=\,\,0.75\text{ }cm\]
C) \[{{l}_{s}}=\,\,0.25\text{ }cm,\,\,\,{{l}_{c}}\,=\,\,0.75\text{ }cm\]
D) \[{{l}_{s}}=\,\,0.75\text{ }cm,\,\,\,{{l}_{c}}\,=\,\,0.25\text{ }cm\]
Correct Answer: A
Solution :
\[l\propto \frac{1}{Y}\,\Rightarrow \,\frac{{{Y}_{s}}}{{{Y}_{c}}}\,=\,\frac{{{l}_{c}}}{{{l}_{s}}}\,\Rightarrow \,\frac{{{l}_{c}}}{{{l}_{s}}}\,\,=\,\,\frac{2\times {{10}^{11}}}{1.2\times {{10}^{11}}}\,\,=\,\,\frac{5}{3}\] ? (i) Also, \[{{l}_{c}}-{{l}_{s}}\,\,=\,\,0.5\] ? (ii) On solving (i) and (ii) \[4 = 1.25 cm\] and \[{{l}_{s}}=\,\,0.75\text{ }cm.\]
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