question_answer

If the ratio of lengths, radii and Youngs 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 :

Free body diagram of the two blocks are Given, \[\frac{{{l}_{1}}}{{{l}_{2}}}=a,\,\frac{{{r}_{1}}}{{{r}_{2}}}=b,\,\frac{{{Y}_{1}}}{{{Y}_{2}}}=c\] Let Youngs modulus of steel is Y^ and of brass is \[{{Y}_{2}}\]. \[\therefore \] \[{{Y}_{1}}=\frac{{{F}_{1}}.{{l}_{1}}}{{{A}_{1}}.\,\Delta {{l}_{2}}}\]            ??(i) and \[{{Y}_{2}}=\frac{{{F}_{2}}.{{l}_{2}}}{{{A}_{2}}\,\Delta {{l}_{2}}}\] ??.(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{\frac{{{F}_{1}}.{{l}_{1}}}{{{A}_{1}}.\Delta {{l}_{1}}}}{\frac{{{F}_{2}}.{{l}_{2}}}{{{A}_{2}}.\Delta {{l}_{2}}}}\] or \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{{{F}_{1}}.{{A}_{2}}.{{l}_{1}}.\Delta {{l}_{2}}}{{{F}_{2}}.{{A}_{1}}.{{l}_{2}}.\Delta {{l}_{1}}}\]          ?..(iii) Force on steel wire from free body diagram \[T={{F}_{1}}=(2g)\] newton Force on brass wire from free body diagram \[{{F}_{2}}=T=T+2g=(4g)\] newton Now, putting the value of \[{{F}_{1}},{{F}_{2}},\]in Eq. (iii), we get \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\left( \frac{2g}{4g} \right).\left( \frac{\pi r_{2}^{2}}{\pi r_{1}^{2}} \right).\left[ \frac{{{l}_{1}}}{{{l}_{2}}} \right].\left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)\] or \[c=\frac{1}{2}\left( \frac{1}{{{b}^{2}}} \right).a\left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)\] or \[\frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}=\left( \frac{a}{2{{b}^{2}}c} \right)\]