A) \[\frac{{{b}^{2}}a}{2c}\]
B) \[\frac{bc}{2{{a}^{2}}}\]
C) \[\frac{b{{a}^{2}}}{2c}\]
D) \[\frac{a}{2{{b}^{2}}c}\]
Correct Answer: D
Solution :
Given, \[\frac{{{l}_{1}}}{{{l}_{2}}}=a,\,\,\frac{{{r}_{1}}}{{{r}_{2}}}=b,\,\,\frac{{{Y}_{1}}}{{{Y}_{2}}}=c\] Let Young?s modulus of steel be \[{{Y}_{1}}\] and that of brass be \[{{Y}_{2}}\] \[\therefore \,\,\,{{Y}_{1}}=\frac{{{F}_{1}}{{l}_{1}}}{{{A}_{1}}\Delta {{l}_{1}}}\] ?. (i) and \[{{Y}_{2}}=\,\,\frac{{{F}_{2}}{{l}_{2}}}{{{A}_{2}}\Delta {{l}_{2}}}\] ?. (ii) Dividing equation (i) by equation (ii), we get \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{{{F}_{1}}{{A}_{2}}{{l}_{1}}\Delta {{l}_{2}}}{{{F}_{2}}{{A}_{1}}{{l}_{2}}\Delta {{l}_{1}}}\] ? (iii) Force on steel wire from free body diagram \[T = {{F}_{1}} = (2g) Newton\] Force on brass wire from free body diagram \[{{\operatorname{F}}_{2}}={{T}_{1}}=T+2g=4g\,\,Newton\] Now putting the value of \[{{F}_{1}},\,\,{{F}_{2}}\] in Equation (iii), we get \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\left( \frac{2g}{4g} \right)\left( \frac{\pi r_{2}^{2}}{\pi r_{1}^{2}} \right).\,\left[ \frac{{{l}_{1}}}{{{l}_{2}}} \right].\left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)=\frac{1}{2}\left( \frac{1}{{{b}^{2}}} \right)\,\,.\,a\left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)\]You need to login to perform this action.
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