(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V}{L}$.Since $V = iR$ and resistance $

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(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V}{L}$.Since $V = iR$ and resistance $R = \rho \frac{L}{A}$,we can substitute these into the expression:$x = \frac{iR}{L} = \frac{i (\rho L / A)}{L} = \frac{i \rho}{A}$.Given:$i = 0.2 \, A$$\rho = 40 \times 10^{-8} \, \Omega \cdot m$$A = 8 \times 10^{-6} \, m^2$Substituting the values:$x = \frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}}$$x = \frac{8 \times 10^{-8}}{8 \times 10^{-6}} = 1 \times 10^{-2} \, V/m = 10^{-2} \, V/m$.

Class 12 Physics Current Electricity Potentiometer

Medium

The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

A
$10^{-2} \, V/m$
B
$10^{-1} \, V/m$
C
$3.2 \times 10^{-2} \, V/m$
D
$1 \, V/m$

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Current Electricity

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Potentiometer

Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.

Medium

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In a potentiometer arrangement,a cell of $emf$ $1.25\,V$ gives a balance point at $35.0\,cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\,cm,$ the $emf$ of the second cell is ............... $V$.

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To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

DifficultMHT CET 2021

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In a potentiometer experiment for the determination of the internal resistance of a cell,when an external resistance of $R$ is connected in parallel to the cell,the balancing length decreases by $10 \%$. The internal resistance of the cell is

MediumTS EAMCET 2025

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$A$ potentiometer has a uniform wire of length $5 \,m$. $A$ battery of emf $10 \,V$ and negligible internal resistance is connected between its ends. $A$ secondary cell connected to the circuit gives a balancing length at $200 \,cm$. The emf of the secondary cell is: (in $\,V$)

MediumKCET 2020

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The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

Similar Questions

Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.

In a potentiometer arrangement,a cell of $emf$ $1.25\,V$ gives a balance point at $35.0\,cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\,cm,$ the $emf$ of the second cell is ............... $V$.

To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

In a potentiometer experiment for the determination of the internal resistance of a cell,when an external resistance of $R$ is connected in parallel to the cell,the balancing length decreases by $10 \%$. The internal resistance of the cell is

$A$ potentiometer has a uniform wire of length $5 \,m$. $A$ battery of emf $10 \,V$ and negligible internal resistance is connected between its ends. $A$ secondary cell connected to the circuit gives a balancing length at $200 \,cm$. The emf of the secondary cell is: (in $\,V$)

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