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(A) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$,where $\rho$ is the resistivity,$l$ is the length,and $r$

Vedclass · Accepted Answer

$0.036$

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(A) The resistance $R$ of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$,where $ho$ is the resistivity,$l$ is the length,and $r$ is the radius.For the first wire: $l_1 = 1 	ext{ cm}$,$r_1 = 1 	ext{ mm}$,$R_1 = 3 	imes 10^{-3} \Omega$.The ratio of resistances for two wires of the same material is given by $\frac{R_2}{R_1} = \frac{l_2}{l_1} 	imes \left(\frac{r_1}{r_2}ight)^2$.Substituting the given values: $l_1 = 1 	ext{ cm}$,$l_2 = 3 	ext{ cm}$,$r_1 = 1 	ext{ mm}$,$r_2 = 0.5 	ext{ mm}$.$\frac{R_2}{3 	imes 10^{-3}} = \left(\frac{3}{1}ight) 	imes \left(\frac{1}{0.5}ight)^2$.$\frac{R_2}{3 	imes 10^{-3}} = 3 	imes (2)^2 = 3 	imes 4 = 12$.$R_2 = 12 	imes 3 	imes 10^{-3} = 36 	imes 10^{-3} \Omega = 0.036 \Omega$.

$A$ metal wire of $1 \text{ cm}$ length and $1 \text{ mm}$ radius has a resistance of $3 \times 10^{-3} \Omega$. If a wire of the same metal of length $3 \text{ cm}$ and radius $0.5 \text{ mm}$ is drawn,what is the resistance of the new wire (in $\Omega$)?

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