Question

A certain wire of length l and radius r has resistance R . if the length of the wire halves the previous value then the radius is doubled, the new resistance will be-
(A) R/4
(B) R/3
(C) 4R
(D) R/8

Answer 1

Hint
Specific resistance is a characteristic of the material being employed and is measured for unit length of material with unit cross section area.If you use a wire if twice the thickness of another, then it should have half the resistance ergo twice the current carrying capacity.

Complete step by step answer
As we know,
 $ R = \dfrac{{\rho L}}{A} $ ......(1)
If the wire having the radius is double than the length was remaining unchanged
Let, the previous value of radius is r and now the radius is 2r
So, now the area of the certain wire is $ A' = \pi {(2r)^2} $
So, the resistance will be $ R' = \dfrac{{\rho l}}{{\pi {{(2r)}^2}}} $
So, $ R' = \dfrac{R}{4} $
New resistance is $ \dfrac{R}{4} $ .
Option (A) is correct.

Note
Resistance is directly proportional to the length of the wire, and inversely proportional to the cross sectional area of the wire.Doubling the length will double the resistance, but the wire also must get thinner as it is stretched, because it will contain the same amount of metal in twice the length.