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Question: A metal wire has diameter of 0.25mm and electrical resistivity of \(0.8\times {{10}^{-8}}\Omega m\) ...

A metal wire has diameter of 0.25mm and electrical resistivity of 0.8×10−8Ωm0.8\times {{10}^{-8}}\Omega m
(a) What will be the length of this wire to make a resistance 5Ω5\Omega ?
(b) How much will the resistance change, if the diameter of the wire is doubled?

Explanation

Solution

As a very first step, one could read the question well and hence note down the given values from it. Now recall the expression for electrical resistance in terms of resistivity, area and length which is used to solve both sub questions. Now substitute accordingly and hence find the answer.

Formula used:
Resistance,
R=ρlA=ρlπr2R=\dfrac{\rho l}{A}=\dfrac{\rho l}{\pi {{r}^{2}}}

Complete step by step solution:
In the question, we are given a metal wire of diameter of 0.25mm and an electrical resistivity of ρ=0.8×10−8Ωm\rho =0.8\times {{10}^{-8}}\Omega m. This information is followed by two sub-questions.
(a) Here, we are supposed to find the length of this wire that would give a resistance of 5Ω5\Omega .
Radius of the metal wire would be,
r=0.252mm=0.125×10−3mr=\dfrac{0.25}{2}mm=0.125\times {{10}^{-3}}m
So, we have its area of cross section to be,
A=πr2=π(0.125×10−3)2=0.05×10−6m2A=\pi {{r}^{2}}=\pi {{\left( 0.125\times {{10}^{-3}} \right)}^{2}}=0.05\times {{10}^{-6}}{{m}^{2}}
Now we have the expression for resistance in terms of area and length given by,
R=ρlA=ρlπr2R=\dfrac{\rho l}{A}=\dfrac{\rho l}{\pi {{r}^{2}}}………………………………………………… (1)
Substituting the given values,
l=5×0.05×10−60.8×10−8l=\dfrac{5\times 0.05\times {{10}^{-6}}}{0.8\times {{10}^{-8}}}
∴l=31.25m\therefore l=31.25m
Therefore, we found the required length to be 31.25m.
(b) Here, we have to find the change in resistance, for the diameter of the wire being doubled.
From the equation (1), we see the inverse relationship of resistance with the area of cross section. So, on doubling the diameter doubles the radius and the new resistance would be,
R′=ρlπr′2=ρlπ(2r)2=ρl4πr2R'=\dfrac{\rho l}{\pi r{{'}^{2}}}=\dfrac{\rho l}{\pi {{\left( 2r \right)}^{2}}}=\dfrac{\rho l}{4\pi {{r}^{2}}}
∴R′=R4\therefore R'=\dfrac{R}{4}
Therefore, we found the value of new resistance to be 14\dfrac{1}{4}times that of the initial value.

Note: We could define the resistivity of a material as the measure of the resistance of a given size of this material to electrical conduction. Another name given to this property is specific electrical resistance or may be volume resistivity. However, the ‘resistivity’ is the term that is used more widely than the other two.