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(B) Given: Ratio of lengths $l_1: l_2 = 5: 2$, ratio of diameters $d_1: d_2 = 4: 3$, and ratio of forces $F_1: F_2 = 4: 5$.Since $A = \pi r^2 = \pi (d

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$\frac{63}{88}$

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(B) Given: Ratio of lengths $l_1: l_2 = 5: 2$, ratio of diameters $d_1: d_2 = 4: 3$, and ratio of forces $F_1: F_2 = 4: 5$.Since $A = \pi r^2 = \pi (d/2)^2$, the ratio of areas is $A_1: A_2 = d_1^2: d_2^2 = 4^2: 3^2 = 16: 9$.Young's modulus is given by $Y = \frac{FL}{A \Delta l}$, so the increase in length is $\Delta l = \frac{FL}{AY}$.Therefore, the ratio of increase in length is $\frac{\Delta l_1}{\Delta l_2} = \frac{F_1}{F_2} 	imes \frac{l_1}{l_2} 	imes \frac{A_2}{A_1} 	imes \frac{Y_2}{Y_1}$.Substituting the values: $\frac{\Delta l_1}{\Delta l_2} = \left(\frac{4}{5}ight) 	imes \left(\frac{5}{2}ight) 	imes \left(\frac{9}{16}ight) 	imes \left(\frac{0.7 	imes 10^{11}}{1.1 	imes 10^{11}}ight)$.$\frac{\Delta l_1}{\Delta l_2} = 2 	imes \frac{9}{16} 	imes \frac{7}{11} = \frac{18}{16} 	imes \frac{7}{11} = \frac{9}{8} 	imes \frac{7}{11} = \frac{63}{88}$.Thus, the ratio is $\frac{63}{88}$.

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