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(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the specific resistance (resistivity),$l$ is the lengt

Vedclass · Accepted Answer

$5:8$

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(C) The resistance $R$ of a wire is given by the formula $R = ho \frac{l}{A}$,where $ho$ is the specific resistance (resistivity),$l$ is the length,and $A$ is the cross-sectional area.Given the ratios:$\frac{ho_1}{ho_2} = \frac{2}{3}$$\frac{l_1}{l_2} = \frac{3}{4}$$\frac{A_1}{A_2} = \frac{4}{5}$The ratio of their resistances is $\frac{R_1}{R_2} = \frac{ho_1}{ho_2} 	imes \frac{l_1}{l_2} 	imes \frac{A_2}{A_1}$.Substituting the given values:$\frac{R_1}{R_2} = \frac{2}{3} 	imes \frac{3}{4} 	imes \frac{5}{4} = \frac{2 	imes 3 	imes 5}{3 	imes 4 	imes 4} = \frac{30}{48} = \frac{5}{8}$.Thus,the ratio of their resistances is $5:8$.

Two wires are made of two different materials whose specific resistances are in the ratio $2:3$,lengths are in the ratio $3:4$,and cross-sectional areas are in the ratio $4:5$. The ratio of their resistances is:

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