The specific resistance (resistivity) of mang | vedclass

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(A) The resistance $R$ of a conductor is given by the formula $R = \frac{\rho l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the

Vedclass · Accepted Answer

$10^{-6} \, \Omega$

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(A) The resistance $R$ of a conductor is given by the formula $R = \frac{ho l}{A}$,where $ho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.Given: $ho = 50 	imes 10^{-8} \, \Omega \cdot m$,$l = 50 \, cm = 0.5 \, m$,and for a cube,the area $A = l^2 = (0.5 \, m)^2 = 0.25 \, m^2$.Substituting these values into the formula:$R = \frac{50 	imes 10^{-8} 	imes 0.5}{0.25}$$R = \frac{25 	imes 10^{-8}}{0.25}$$R = 100 	imes 10^{-8} \, \Omega = 10^{-6} \, \Omega$.Therefore,the correct option is $A$.

The specific resistance (resistivity) of manganin is $50 \times 10^{-8} \, \Omega \cdot m$. The resistance of a cube of length $50 \, cm$ will be:

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