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 :

Let Youngs modulus of steel and brass are \[{{Y}_{1}}\] and \[{{Y}_{2}}\] respectively. \[\therefore \]  \[{{Y}_{1}}=\frac{{{F}_{1}}}{{{A}_{1}}}\cdot \frac{{{l}_{1}}}{\Delta {{l}_{1}}}\]and \[{{Y}_{2}}=\frac{{{F}_{2}}}{{{A}_{2}}}\cdot \frac{{{l}_{2}}}{\Delta {{l}_{2}}}\] Now, free body diagram of the two blocks are Force on steel wire \[T={{F}_{1}}=2g\,N.\] Force on brass wire\[{{F}_{2}}=T=T+2g=4g\,N\] \[\frac{{{Y}_{1}}}{{{Y}_{2}}}=\,\,\,\left( \frac{2g}{4g} \right)\cdot \left( \frac{\pi r_{2}^{2}}{\pi r_{1}^{2}} \right)\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)\cdot \left( \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}} \right)\] \[c=\frac{1}{2}\cdot \frac{1}{{{b}^{2}}}\cdot a\cdot \frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}\] or  \[\frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}=\frac{a}{2{{b}^{2}}c}\]