Question

The diameter of a wire is reduced to one-fifth of its original value by stretching it. Its internal resistance is R, what would be its resistance after reduction of the diameter?
(A) $\dfrac{R}{625}$
(B) $\dfrac{R}{25}$
(C) 25R
(D) 625R

Answer 1

Hint: The resistance of a wire is directly proportional to the length of the wire and inversely proportional to the area of cross-section of the wire. In stretching the wire, the volume of the wire remains the same and the length and diameter change.
Formula used:
The resistance of the wire is given as:
$R =\rho \dfrac{L}{A}$ .
Here $\rho$ is the resistivity of the material that depends on the nature and temperature of the material, L is for the length of the wire and A is for the area of cross-section.

Complete answer:
We multiply the numerator and denominator of the resistance formula by A, we get:
$R =\rho \dfrac{LA}{A^2} = \rho \dfrac{V}{A^2}$
Now, we have modified the formula because all parameters in this formula remain constant now except for area (or diameter).
As we substitute the formula for area in terms of diameter,
$R = \rho \dfrac{V}{ (\pi (d/2)^2)^2 }$
Now, we are given that upon stretching, the diameter reduces to one-fifth of the initial value.
$R' = \rho \dfrac{V}{ (\pi (d/(5)2)^2)^2 }$
Now, if we use the previous relation of R here,
$R' = 625 R$ .

Therefore, the correct answer is option (D).

Additional Information:
The resistivity of any material happens to be an important intrinsic property of the material. On calculating the resistivity, one can determine the nature of electrical conductivity the material offers.

Note:
The formula has to contain all the constant parameters like resistivity and volume. If one uses the formula for resistance without making these changes then one will land on the wrong answer for sure in that case. This is so because, when we are stretching the wire then directly two parameters are changing, one the length and other, the area.