Correct Answer - Option 3 : 100 Ω
Concept:
Resistance,
\(R=\frac{ρ\ l}{A}\) .......(1)
Where ρ = resistivity of the conductor (Ω m)
l = length of conductor (m)
A = cross-section area of conductor (m2)
A = π r2 .........(2)
Where r is the radius of the cross-section of the conductor.
Explanation:
From equations (1) and (2),
\(R\propto\frac{1}{r^2}\) as length 'l' and resistivity 'ρ' is constant.
\(\large{\frac{R_1}{R_2}=\frac{r_2^2}{r_1^2}}\)
Diameter is doubled means radius is doubled.
Given, R1 = 400 Ω , r2 = 2 r1
Where,
R1, r1 is the old resistance and old radius of the wire.
R2, r2 is the new resistance and new radius of the wire.
\(\large{\frac{400}{R_2}=\frac{(2r_1)^2}{(r_1)^2}=4}\)
R2 = 100 Ω
Therefore, new resistance R2 = 100 Ω
