Question
If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are $$a,b$$ and $$c$$ respectively, then the corresponding ratio of increase in their lengths is:
If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are $$a,b$$ and $$c$$ respectively, then the corresponding ratio of increase in their lengths is:
A.
$$\frac{{3c}}{{2a{b^2}}}$$
B.
$$\frac{{2{a^2}c}}{b}$$
C.
$$\frac{{3a}}{{2{b^2}c}}$$
D.
$$\frac{{2ac}}{{{b^2}}}$$
Answer :
$$\frac{{3a}}{{2{b^2}c}}$$
Solution :
According to questions,
$$\eqalign{ & \frac{{{\ell _s}}}{{{\ell _b}}} = a,\frac{{{r_s}}}{{{r_b}}} = b,\frac{{{y_s}}}{{{y_b}}} = c,\frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = ? \cr & {\text{As,}}\,\,y = \frac{{F\ell }}{{A\Delta \ell }} \Rightarrow \Delta \ell = \frac{{F\ell }}{{Ay}} \cr & \Delta {\ell _s} = \frac{{3mg{\ell _s}}}{{\pi r_s^2.{y_s}}}\,\,\left[ {\because {F_s} = \left( {M + 2M} \right)g} \right] \cr & \Delta {\ell _b} = \frac{{2Mg{\ell _b}}}{{\pi r_b^2.{y_b}}}\,\,\left[ {\because {F_b} = 2Mg} \right] \cr & \therefore \frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = \frac{{\frac{{3Mg{\ell _s}}}{{\pi r_s^2.{y_s}}}}}{{\frac{{2Mg.{\ell _b}}}{{\pi r_b^2.{y_b}}}}} = \frac{{3a}}{{2{b^2}c}} \cr} $$
According to questions,
$$\eqalign{ & \frac{{{\ell _s}}}{{{\ell _b}}} = a,\frac{{{r_s}}}{{{r_b}}} = b,\frac{{{y_s}}}{{{y_b}}} = c,\frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = ? \cr & {\text{As,}}\,\,y = \frac{{F\ell }}{{A\Delta \ell }} \Rightarrow \Delta \ell = \frac{{F\ell }}{{Ay}} \cr & \Delta {\ell _s} = \frac{{3mg{\ell _s}}}{{\pi r_s^2.{y_s}}}\,\,\left[ {\because {F_s} = \left( {M + 2M} \right)g} \right] \cr & \Delta {\ell _b} = \frac{{2Mg{\ell _b}}}{{\pi r_b^2.{y_b}}}\,\,\left[ {\because {F_b} = 2Mg} \right] \cr & \therefore \frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = \frac{{\frac{{3Mg{\ell _s}}}{{\pi r_s^2.{y_s}}}}}{{\frac{{2Mg.{\ell _b}}}{{\pi r_b^2.{y_b}}}}} = \frac{{3a}}{{2{b^2}c}} \cr} $$
