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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
$\frac{{3c}}{{2a{b^2}}}$
$\frac{{2{a^2}c}}{b}$
$\frac{{3a}}{{2{b^2}c}}$
$\frac{{2ac}}{{{b^2}}}$
Solution
According to questions,
$\frac{{{\ell _s}}}{{{\ell _b}}} = a,\frac{{{r_s}}}{{{r_b}}} = b,\frac{{{y_s}}}{{{y_b}}} = c,\frac{{\Delta \ell s}}{{\Delta {\ell _b}}} = ?$
$As,y = \frac{{F\ell }}{{A\Delta \ell }} \Rightarrow \Delta \ell = \frac{{F\ell }}{{Ay}}$
$\Delta {\ell _s} = \frac{{3mg{\ell _s}}}{{\pi _s^2.{y_s}}}\left[ {{F_s} = \left( {M + 2M} \right)g} \right]$
$\Delta {\ell _b} = \frac{{2Mg{\ell _b}}}{{\pi r_b^2{y_b}}}\left[ {{F_b} = 2Mg} \right]$
$\therefore \frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = \frac{{\frac{{3Mg{\ell _s}}}{{\pi r_s^2.{y_s}}}}}{{\frac{{2Mg.{\ell _b}}}{{\pi {b^2}.{y_b}}}}} = \frac{{3a}}{{2{b^2}C}}$
