Two wire of the same meta have same length, but their cross-sections are in the rati `3:1` . They are joined in series. The resistance of thicker wire is `10Omega` . The total resistance of the combination will be

Options

A. `20Omega `
B. `30 Omega `
C. `40 Omega `
D. `50 Omega `

Answer

C

Explanation

`(A_(1 ))/(A_(2)) =(3)/(1) implies R =( rho l)/( A) so (R _(1) )/(R_(2)) =(A _(2) )/(A_(1))=(1)/(3) `
`R_(1) =10 Omega ,R_(2) = 3R_(1)=30 Omega `
in series `R_(eq ) =R _(1) +R_(2) =10+30=40 Omega `