Question

A certain wire has resistance R. The resistance of another wire identical with the first except having twice its diameter is:
A) 2R
B) 0.25R
C) 4R
D) 0.5R

Answer 1

Hint: Here we must compare the resistances of two different wires made of the same material and identical to each other, but the only difference is their diameter. By using the relation between resistance, resistivity, length of the wire and area of cross section we can solve this question.

Complete step by step answer:
Let us write all the given data, the resistance of the wire is R. we must find the resistance of the wire which has twice its diameter.
First let us consider the first wire having resistance R,
$R=\rho\dfrac{L}{A}$ -----(i)
Where R is the resistance, \rho is the resistivity, L is the length of the wire and A is the cross-sectional area of the wire. Area of cross section $A=\pi r^2$ r is the radius of the wire.
Since this is to be compared with another wire with diameter of wire, hence we write radius in terms of diameter. i.e. d=2r Area can be written as follows,
$A=\pi\left(\dfrac{d}{2}\right)^2=\dfrac{\pi d^2}{4}$ Using the same in equation (i)
$R=\dfrac{\rho L}{\dfrac{\pi d^2}{4}}=\dfrac{4\rho L}{\pi d^2}$ ---------(ii)
For another wire having diameter twice the first one.
$R_1=\dfrac{4\rho L}{\pi\left(2d\right)^2} $
$\implies R_1=\dfrac{4\rho L}{\pi{4d}^2}=\dfrac{\rho L}{\pi d^2}$ -------- (iii)
From equation (ii) we have $R=\dfrac{4\rho L}{\pi d^2} $
But from equation (iii) $R_1=\dfrac{\rho L}{\pi d^2} $
By using (ii) and (iii) $R=4R_1$
Hence $R_1=\dfrac{R}{4}=0.25R$
We conclude that the resistance of the second wire is one fourth of the first wire.
Option (B) is the correct answer.

So, the correct answer is “Option B”.

Note:
The resistivity is the property is the material which is constant for each material.
Even though any other factors like length, area changes the resistivity will be the same.