The specific resistance (resistivity) of mate | vedclass

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(D) Given: $\rho_B = 2\rho_A$ and $d_B = 2d_A$,which implies $r_B = 2r_A$.Since the resistances are equal,$R_A = R_B$.Using the formula for resistance

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$2$

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(D) Given: $ho_B = 2ho_A$ and $d_B = 2d_A$,which implies $r_B = 2r_A$.Since the resistances are equal,$R_A = R_B$.Using the formula for resistance $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$,we have:$\frac{ho_A l_A}{\pi r_A^2} = \frac{ho_B l_B}{\pi r_B^2}$Rearranging for the ratio of lengths:$\frac{l_B}{l_A} = \frac{ho_A}{ho_B} 	imes \left( \frac{r_B}{r_A} ight)^2$Substituting the given values:$\frac{l_B}{l_A} = \frac{ho_A}{2ho_A} 	imes \left( \frac{2r_A}{r_A} ight)^2$$\frac{l_B}{l_A} = \frac{1}{2} 	imes (2)^2 = \frac{1}{2} 	imes 4 = 2$.Thus,the ratio $l_B/l_A$ is $2$.

The specific resistance (resistivity) of material $B$ is twice that of material $A$. The diameter of a circular wire made of $B$ is twice that of a wire made of $A$. If the resistances of both wires are equal,then the ratio of their lengths $l_B/l_A$ will be .......

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