The radius of a conducting wire is doubled .what will be the ratio of its new specific resistance to the old one?

The equation for the resistance R is: R=ρ*(l/A), where, ρ is electrical resistivity, l is the length of the conductor, and A is the surface area.

The initial surface area is:

A=r²π, then when we double the radius we get:
 
A₁=(2*r)²π=4*r²π=4*A

Initial resistance is: R=ρ*(l/A).

When we double the radius, resistance is: R₁=ρ*{ l / (4*A) }

The ratio of the new resistance to the old one:

R₁/R=[ρ*(l/A)] / [ ρ* { l / (4*A) } ] = ρ, l and A cancel out and we get:

R₁/R=(1/1)/(1/4)=4/1 

The correct answer is c. hope this

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