If the ratio of lengths, radii and Young' | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{{3a}}{{2{b^2}c}}$

Vedclass · Answer

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

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

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