(C) 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 $R = \fra

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(C) 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 $R = \frac{\rho L}{A}$, we can substitute these into the expression for the potential gradient:$x = \frac{I \cdot (\rho L / A)}{L} = \frac{I \rho}{A}$.Given values are:Current $(I) = 0.1 \, A$,Resistivity $(\rho) = 10^{-7} \, \Omega \cdot m$,Area $(A) = 10^{-6} \, m^2$.Substituting these values into the formula:$x = \frac{0.1 \times 10^{-7}}{10^{-6}} = \frac{10^{-1} \times 10^{-7}}{10^{-6}} = \frac{10^{-8}}{10^{-6}} = 10^{-2} \, V/m$.

Class 12 Physics Current Electricity Potentiometer

Medium

$A$ potentiometer has a uniform potential gradient. The specific resistance (resistivity) of the material of the potentiometer wire is $10^{-7} \, \Omega \cdot m$, the current passing through it is $0.1 \, A$, and the cross-sectional area of the wire is $10^{-6} \, m^2$. The potential gradient along the potentiometer wire is:

AIPMT 2001 All AIPMT

A
$10^{-4} \, V/m$
B
$10^{-6} \, V/m$
C
$10^{-2} \, V/m$
D
$10^{-8} \, V/m$

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Current Electricity

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Potentiometer

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In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is ............... $m$.

Medium

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As shown in the figure,a potentiometer wire of resistance $20\,\Omega$ and length $300\,cm$ is connected with a resistance box ($R$.$B$.) and a standard cell of emf $4\,V$. For a resistance '$R$' of the resistance box introduced into the circuit,the null point for a cell of $20\,mV$ is found to be $60\,cm$. The value of '$R$' is $.....\Omega$

MediumJEE MAIN 2022

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$A$ potentiometer wire has a length of $5 \, m$ and a resistance of $16 \, \Omega$. The driving cell has an e.m.f. of $5 \, V$ and an internal resistance of $4 \, \Omega$. When two cells of e.m.f.s $1.3 \, V$ and $1.1 \, V$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively:

DifficultMHT CET 2023

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$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

Medium

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$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

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In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is ............... $m$.

$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

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