Explanation

(d) Given, $\frac{l_{1}}{l_{2}} = a,\frac{r_{1}}{r_{2}} = b,\frac{Y_{1}}{Y_{2}} = c$
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Let Young’s modulus of steel be $Y_{1}$, and that of brass be $Y_{2}$
$\therefore Y_{1} = \frac{F_{1}l_{1}}{A_{1}\mathrm{\Delta}l_{1}}$ …(i)
and $Y_{2} = \frac{F_{2}l_{2}}{A_{2}\mathrm{\Delta}l_{2}}$ …(ii)
Dividing Equation (i) by Equation (ii), we get
$\frac{Y_{1}}{Y_{2}} = \frac{F_{1}.A_{2}.l_{1}.\mathrm{\Delta}l_{2}}{F_{2}.A_{1}.l_{2}.\mathrm{\Delta}l_{1}}$ …(iii)
Force on steel wire from free body diagram
$T = F_{1} = (2g)$ Newton
Force on brass wire from free body diagram
$F_{2} = T_{1}^{''} = T + 2g = 4g$ Newton
Now, putting the value of $F_{1},F_{2},$ in Equation (iii), we get
$\frac{Y_{1}}{Y_{2}} = \left( \frac{2g}{4g} \right).\left( \frac{{\pi r}_{2}^{2}}{\pi r_{1}^{2}} \right).\left\lbrack \frac{l_{1}}{l_{2}} \right\rbrack.\left( \frac{\mathrm{\Delta}l_{2}}{\mathrm{\Delta}l_{1}} \right) = \frac{1}{2}\left( \frac{1}{b_{2}} \right).a\left( \frac{\mathrm{\Delta}l_{2}}{\mathrm{\Delta}l_{1}} \right)$