A) \[1.5:2\]
B) \[1.8:2\]
C) \[1.5:1\]
D) \[1.8:1\]
Correct Answer: D
Solution :
As given for steel wire \[{{A}_{1}}=3\times {{10}^{-5}}{{m}^{2}},\,\,{{l}_{1}}=4.7\,m,\,\Delta {{l}_{1}}=\Delta l,\,{{F}_{2}}=F\] For copper wire \[{{A}_{2}}=4\times {{10}^{-5}}\,{{m}^{2}},\,{{l}_{2}}=3.5\,m,\,\Delta {{L}_{2}}=\Delta l,\,{{F}_{2}}=F\] Let \[{{Y}_{1}}\] and \[{{Y}_{2}}\] be the Young?s modulus of steel wire and copper wire, respectively So, \[{{Y}_{1}}=\frac{{{F}_{1}}}{{{A}_{1}}}\times \frac{{{l}_{1}}}{\Delta {{l}_{1}}}\,=\frac{F}{3\times {{10}^{-5}}}\,\times \frac{4.7}{\Delta l}\] and \[{{Y}_{2}}=\frac{{{F}_{2}}}{{{A}_{2}}}\,\times \frac{{{l}_{2}}}{\Delta {{l}_{2}}}\,=\frac{F}{4\times {{10}^{-5}}}\,\times \frac{3.5}{\Delta l}\] \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{4.7\times 10\times {{10}^{-5}}}{3.5\times 3\times {{10}^{-5}}}=1.8\] So, \[{{Y}_{1}}:{{Y}_{2}}=1.8:1\]
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