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Question: A uniform wire of length l and radius r has a resistance of \[100\Omega \]. It is recast into a wire...

A uniform wire of length l and radius r has a resistance of 100Ω100\Omega . It is recast into a wire of radius r2\dfrac{r}{2}. The resistance of new 2 wire will be :
(A) 100Ω100\Omega
(B) 1600Ω1600\Omega
(C) 200Ω200\Omega
(D) 400Ω400\Omega

Explanation

Solution

Hint Use the resistance formula which is given byR=ρlAR = \dfrac{{\rho l}}{A}. Consider both the cases where radius is r and radius of r2\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 100Ω100\Omega . Thus, the wire is said to have a resistivity ρA{\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=ρAlAR = \dfrac{{{\rho _A}l}}{A}
R=ρAlπr2\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
R2=ρAlπr22\Rightarrow {R_2} = \dfrac{{{\rho _A}l}}{{\pi {r_2}^2}}
Now here, r2=r2{r_2} = \dfrac{r}{2}
R2=ρAlπr2/4\Rightarrow {R_2} = \dfrac{{{\rho _A}l}}{{\pi {r^2}/4}}-------- (2)
Dividing equation (1) by equation (2):
RR2=ρAlπr2ρAlπr2/4\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
RR2=1πr21πr2/4\Rightarrow \dfrac{R}{{{R_2}}} = \dfrac{{\dfrac{1}{{\pi {r^2}}}}}{{\dfrac{1}{{\pi {r^2}/4}}}}
RR2=πr24πr2\Rightarrow \dfrac{R}{{{R_2}}} = \dfrac{{\pi {r^2}}}{{4\pi {r^2}}}
Cancelling out the common terms, we get
100R2=14\Rightarrow \dfrac{{100}}{{{R_2}}} = \dfrac{1}{4}
R2=400Ω\Rightarrow {R_2} = 400\Omega
Hence, for a given material with resistance of100Ω100\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.