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

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(B) Young's modulus is given by $Y = \frac{F L}{A \Delta L} = \frac{4 F L}{\pi D^2 \Delta L}$,where $F$ is the force,$L$ is the length,$A$ is the cros

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$\frac{5q}{3p^2r}$

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(B) Young's modulus is given by $Y = \frac{F L}{A \Delta L} = \frac{4 F L}{\pi D^2 \Delta L}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,$D$ is the diameter,and $\Delta L$ is the change in length.Rearranging for the change in length,we get $\Delta L = \frac{4 F L}{\pi D^2 Y}$.For the steel wire,the tension $F_S = (2m + 3m)g = 5mg$. For the brass wire,the tension $F_B = 3mg$.Given ratios: $\frac{D_S}{D_B} = p$,$\frac{L_S}{L_B} = q$,and $\frac{Y_S}{Y_B} = r$.The ratio of the increase in lengths is:$\frac{\Delta L_S}{\Delta L_B} = \left( \frac{F_S}{F_B} \right) \left( \frac{L_S}{L_B} \right) \left( \frac{D_B^2}{D_S^2} \right) \left( \frac{Y_B}{Y_S} \right)$Substituting the given values:$\frac{\Delta L_S}{\Delta L_B} = \left( \frac{5mg}{3mg} \right) (q) \left( \frac{1}{p^2} \right) \left( \frac{1}{r} \right) = \frac{5q}{3p^2r}$.

If the ratio of diameters,lengths and Young's moduli of steel and brass wires shown in the figure are $p, q$ and $r$ respectively,then the corresponding ratio of increase in their lengths would be

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