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

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(B) The Young's modulus $Y$ is defined as $Y = \frac{FL}{A\Delta L} = \frac{4FL}{\pi D^2 \Delta L}$.Rearranging for the change in length,we get $\Delt

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$\frac{7q}{5p^2s}$

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(B) The Young's modulus $Y$ is defined as $Y = \frac{FL}{A\Delta L} = \frac{4FL}{\pi D^2 \Delta L}$.Rearranging for the change in length,we get $\Delta L = \frac{4FL}{\pi D^2 Y}$.Let the subscripts $S$ and $C$ refer to steel and copper wires respectively.The force on the steel wire is $F_S = (5m + 2m)g = 7mg$.The force on the copper wire is $F_C = 5mg$.Given ratios are $\frac{D_S}{D_C} = p$,$\frac{L_S}{L_C} = q$,and $\frac{Y_S}{Y_C} = s$.The ratio of increase in their lengths is $\frac{\Delta L_S}{\Delta L_C} = \left( \frac{F_S}{F_C} \right) \left( \frac{L_S}{L_C} \right) \left( \frac{D_C}{D_S} \right)^2 \left( \frac{Y_C}{Y_S} \right)$.Substituting the given values: $\frac{\Delta L_S}{\Delta L_C} = \left( \frac{7mg}{5mg} \right) (q) \left( \frac{1}{p} \right)^2 \left( \frac{1}{s} \right) = \frac{7q}{5p^2s}$.

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

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