The lengths radii and specific resistance of two conducting wires are each in the ratio of 1:3. If the resistance of the thinner wire is 10 `Omega` then the resistance of the other wire will be

Options

A. 49 `Omega`
B. 20 `Omega`
C. 10 `Omega`
D. 5 `Omega`

Answer

C

Explanation

We know R `=sigma(l)/(A)=(sigmal)/(pir^(2))`
`:." "("R"_(1))/("R"_(2))=(sigma_(1))/(sigma_(2)).(l_(1))/(l_(2)).((r_(2))/(r_(1)))^(2)=(1)/(3)xx(1)/(3).((3)/(1))^(2)=1`
`:." ""R"_(2)="R"_(1)=10Omega`