Question

Determine the ratio of the resistivity of pure water to silver?

Answer 1

Answer:

$3 \cdot 10^{11} : 1$

Explanation:

Silver is a generally excellent conductor, this implies its resistivity is low (from table, we can check the exact worth i.e $\rho_s = 1.6 \cdot 10^{-8} \Omega m$)

Pure water, rather, is an extremely awful conductor, this implies its resistivity is exceptionally high, of order of $k \Omega \cdot m (10^3 \Omega m )$

Indeed, even without knowing the exact estimation of the pure water resistivity, we can appraise the proportion between the pure water resistivity and the silver resistivity by looking at the two orders of magnitude:

$r= \frac{10^3 \Omega m}{10^{-8} \Omega m} \sim 10^{11}$

Therefore, we can say that the correct answer is

$3 \cdot 10^{11} : 1$