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

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(C) From the figure,the force acting on the steel wire is $F_s = (M + 2M)g = 3Mg$,and the force acting on the brass wire is $F_b = 2Mg$.The formula fo

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

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(C) From the figure,the force acting on the steel wire is $F_s = (M + 2M)g = 3Mg$,and the force acting on the brass wire is $F_b = 2Mg$.The formula for the increase in length is $\Delta \ell = \frac{F \ell}{A Y} = \frac{F \ell}{\pi r^2 Y}$.Given ratios: $\frac{\ell_s}{\ell_b} = a$,$\frac{r_s}{r_b} = b$,and $\frac{Y_s}{Y_b} = c$.Now,the ratio of the increase in lengths is:$\frac{\Delta \ell_s}{\Delta \ell_b} = \frac{F_s \ell_s / (\pi r_s^2 Y_s)}{F_b \ell_b / (\pi r_b^2 Y_b)}$Substituting the values:$\frac{\Delta \ell_s}{\Delta \ell_b} = \left( \frac{F_s}{F_b} \right) \left( \frac{\ell_s}{\ell_b} \right) \left( \frac{r_b}{r_s} \right)^2 \left( \frac{Y_b}{Y_s} \right)$$\frac{\Delta \ell_s}{\Delta \ell_b} = \left( \frac{3Mg}{2Mg} \right) \cdot (a) \cdot \left( \frac{1}{b} \right)^2 \cdot \left( \frac{1}{c} \right)$$\frac{\Delta \ell_s}{\Delta \ell_b} = \frac{3}{2} \cdot a \cdot \frac{1}{b^2} \cdot \frac{1}{c} = \frac{3a}{2b^2c}$.

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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