Question

A uniform wire of length l and radius r has a resistance of \[100\Omega \]. It is recast into a wire of radius \[\dfrac{r}{2}\]. The resistance of new 2 wire will be :
(A) \[100\Omega \]
(B) \[1600\Omega \]
(C) \[200\Omega \]
(D) \[400\Omega \]

Answer 1

Hint Use the resistance formula which is given by\[R = \dfrac{{\rho l}}{A}\]. Consider both the cases where radius is r and radius of \[\dfrac{r}{2}\]. Compare case 1 with case2 and find the resistance of the second.

Complete Step By Step Solution
It is given that initially an uniform wire of length l and radius r is having a resistance of \[100\Omega \]. Thus, the wire is said to have a resistivity \[{\rho _A}\]. Electrical resistivity is defined as the electrical property of a material which defines its strength to oppose electric current.

Now according to resistance formula, \[R = \dfrac{{{\rho _A}l}}{A}\]
\[ \Rightarrow R = \dfrac{{{\rho _A}l}}{{\pi {r^2}}}\](Since the wire is said to be cylindrical in shape which have a circular surface as its face)------(1)
 In Question, it is given that the same wire is recast into another wire, which has a radius half its initial radius. This implies, that the resistivity of the material will remain since, resistivity depends upon the material properties and not its physical changes.
 Hence, for the second case, the resistance of the material is given as
\[ \Rightarrow {R_2} = \dfrac{{{\rho _A}l}}{{\pi {r_2}^2}}\]
Now here, \[{r_2} = \dfrac{r}{2}\]
\[ \Rightarrow {R_2} = \dfrac{{{\rho _A}l}}{{\pi {r^2}/4}}\]-------- (2)
 Dividing equation (1) by equation (2):
\[ \Rightarrow \dfrac{R}{{{R_2}}} = \dfrac{{\dfrac{{{\rho _A}l}}{{\pi {r^2}}}}}{{\dfrac{{{\rho _A}l}}{{\pi {r^2}/4}}}}\]
Cancelling out the common terms, we get
\[ \Rightarrow \dfrac{R}{{{R_2}}} = \dfrac{{\dfrac{1}{{\pi {r^2}}}}}{{\dfrac{1}{{\pi {r^2}/4}}}}\]
\[ \Rightarrow \dfrac{R}{{{R_2}}} = \dfrac{{\pi {r^2}}}{{4\pi {r^2}}}\]
Cancelling out the common terms, we get
\[ \Rightarrow \dfrac{{100}}{{{R_2}}} = \dfrac{1}{4}\]
\[ \Rightarrow {R_2} = 400\Omega \]
 Hence, for a given material with resistance of\[100\Omega \], when the material is altered in such a way that its radius is halved, the resistance value increases 4 times.

Thus Option(d) is the right answer for the given question.

Note
The basic difference between resistivity and resistance is that resistivity is a property of the material having specific dimensions, whereas resistance is the physical property of the substance that opposes the flow of current. Both are related but their meaning and application differ.