When the temperature of a conductor increase the ratio of its resistivity and conductivity ......

Options

A. Decreases
B. increases
C. remain constant
D. increases or decreases

Answer

B

Explanation

increases
`(rho)/(sigma ) = (rho)/((1)/(rho)) = rho^(2) " " [ because sigma = (1)/(rho) ] `
Reciprocal of resistivity is called conductivity.
Now `rho = (m)/("ne"^(2) tau)` from the equation, on increase the temperature `tau` decreases.
So `rho ` increases, resultant `rho^(2)` increases