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Question: A potentiometer has uniform potential gradient the specific resistance of the material of the potent...

A potentiometer has uniform potential gradient the specific resistance of the material of the potentiometer wire is 10−7{{10}^{-7}} ohms meter and the current passing through it is 0.1 ampere cross section of the wire is10−6m2{{10}^{-6}}{{m}^{2}}. The potential gradient along the potentiometer wire is:
A.10−6V/m B.10−4V/m C.10−8V/m D.10−2V/m \begin{aligned} & A{{.10}^{-6}}V/m \\\ & B{{.10}^{-4}}V/m \\\ & C{{.10}^{-8}}V/m \\\ & D{{.10}^{-2}}V/m \\\ \end{aligned}

Explanation

Solution

We have been provided with potentiometer and value of the instrument used in potentiometer circuit potential gradient is the product of resistance gradient and current flows through circuit. Now to calculate resistance gradient, use a formula of specific resistance which gives the relation between area of cross section length of wire and resistance.

Formula Used:
Potential gradient is given by,
V′=IR′V'=IR'
Where, V′V' = potential gradient
I = current flow
R′R' = resistance gradient.

Complete step by step answer:
We have given a potentiometer which has uniform potential gradient. The specific resistance of material of potentiometer wire which is denoted by ρ\rho is 10−7{{10}^{-7}}ohms meter.
I.e. resistivity (ρ)=10−7Ωm\left( \rho \right)={{10}^{-7}}\Omega m
The current passing through the potentiometer denoted by I is 0.1 A.
I.e. current (i) = 0.1A
The wire use in circuit has area of cross section denoted by A is 10−6m2{{10}^{-6}}{{m}^{2}}
Area of cross section (A) = 10−6m2{{10}^{-6}}{{m}^{2}}
We know that resistance of conductor is directly proportional to it length and inversely proportional to its area of cross section (A)
R∝l and R∝1A R∝lA R=ρlA \begin{aligned} & R\propto l\text{ and R}\propto \dfrac{1}{A} \\\ & R\propto \dfrac{l}{A} \\\ & R=\dfrac{\rho l}{A} \\\ \end{aligned}
Where ρ\rho is the resistivity of material we know that, resistance gradient is nothing but ratio of resistance and length of wire which is given by,
Resistance gradient, R′=Rl=ρAR'=\dfrac{R}{l}=\dfrac{\rho }{A}
R′=10−710−6=0.1R'=\dfrac{{{10}^{-7}}}{{{10}^{-6}}}=0.1
Therefore, potential gradient is a product of current flow through circuit and resistance gradient which is given by,
V′=iR′ V’=0.1×0.1 V′=10−2V/m \begin{aligned} & V'=iR' \\\ & V’=0.1\times 0.1 \\\ & V'={{10}^{-2}}V/m \\\ \end{aligned}
Hence, potential along the potentiometer is 10−2V/m{{10}^{-2}}V/m.

Therefore option (D) is the correct option.

Note:
ρ\rho is also called as constant of proportionality or specific resistance the resistivity of material is the resistance of wire of unit length and unit area of cross section (potentiometer wire)the resistance of different material (of same dimension) will be different reciprocal of resistivity is conductivity denoted by r.