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(A) The ratio of conductivity $(\sigma)$ to resistivity $(\rho)$ is given by $\frac{\sigma}{\rho}$.Since $\sigma = \frac{1}{\rho}$, we can write the r

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decreases

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(A) The ratio of conductivity $(\sigma)$ to resistivity $(\rho)$ is given by $\frac{\sigma}{\rho}$.Since $\sigma = \frac{1}{\rho}$, we can write the ratio as $\frac{1/\rho}{\rho} = \frac{1}{\rho^2}$.For a conductor, as the temperature increases, the resistivity $(\rho)$ increases due to increased collisions of electrons with the lattice ions.Since $\rho$ is in the denominator and is squared, as $\rho$ increases, the value of $\frac{1}{\rho^2}$ decreases.Therefore, the ratio of conductivity to resistivity decreases.

When the temperature of a conductor increases, the ratio of conductivity and resistivity . . . . . . .

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