question_answer

If the ratio of lengths, radii and Young?s modulus of steel and brass wires shown in the figure are a, b, and c, respectively. The ratio between the increase in lengths of brass and steel wires would be  

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 \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\left( \frac{2g}{4g} \right)\left( \frac{\pi r_{2}^{2}}{\pi r_{1}^{2}} \right).\,\left[ \frac{{{l}_{1}}}{{{l}_{2}}} \right].\left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)=\frac{1}{2}\left( \frac{1}{{{b}^{2}}} \right)\,\,.\,a\left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)\]