(D) The potential gradient $(x)$ is defined as the potential drop per unit length of the wire: $x = \frac{V}{\ell} = \frac{iR}{\ell}$.Since resistance

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(D) The potential gradient $(x)$ is defined as the potential drop per unit length of the wire: $x = \frac{V}{\ell} = \frac{iR}{\ell}$.Since resistance $R = \frac{\rho \ell}{A}$,substituting this into the equation gives: $x = \frac{i (\rho \ell / A)}{\ell} = \frac{i \rho}{A}$.Given values: Current $i = 0.2 \, A$,resistivity $\rho = 4 \times 10^{-7} \, \Omega \cdot m$,and cross-sectional area $A = 8 \times 10^{-7} \, m^2$.Substituting these values into the formula: $x = \frac{0.2 \times 4 \times 10^{-7}}{8 \times 10^{-7}}$.$x = \frac{0.8 \times 10^{-7}}{8 \times 10^{-7}} = 0.1 \, V/m$.

Class 12 Physics Current Electricity Potentiometer

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The current in the primary circuit of a potentiometer is $0.2 \, A$. The specific resistance and cross-section of the potentiometer wire are $4 \times 10^{-7} \, \Omega \cdot m$ and $8 \times 10^{-7} \, m^2$ respectively. The potential gradient will be equal to .............. $V/m$.

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A
$0.2$
B
$1$
C
$0.5$
D
$0.1$

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In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$. The balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_1}{E_2}$ is: (in $: 1$)

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$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-

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The length of a potentiometer wire is $\ell$. $A$ cell of emf $E$ is balanced at a length $\ell/3$ from the positive end of the wire. If the length of the wire is increased by $\ell/2$,at what distance will the same cell give a balanced point?

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$A$ potentiometer has a uniform potential gradient across it. Two cells connected in series $(i)$ to support each other and $(ii)$ to oppose each other are balanced over $6 \ m$ and $2 \ m$ respectively on the potentiometer wire. The ratio of the $e.m.f.$s of the cells is:

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$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire is (in $m$)

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The current in the primary circuit of a potentiometer is $0.2 \, A$. The specific resistance and cross-section of the potentiometer wire are $4 \times 10^{-7} \, \Omega \cdot m$ and $8 \times 10^{-7} \, m^2$ respectively. The potential gradient will be equal to .............. $V/m$.

Similar Questions

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$. The balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_1}{E_2}$ is: (in $: 1$)

$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-

The length of a potentiometer wire is $\ell$. $A$ cell of emf $E$ is balanced at a length $\ell/3$ from the positive end of the wire. If the length of the wire is increased by $\ell/2$,at what distance will the same cell give a balanced point?

$A$ potentiometer has a uniform potential gradient across it. Two cells connected in series $(i)$ to support each other and $(ii)$ to oppose each other are balanced over $6 \ m$ and $2 \ m$ respectively on the potentiometer wire. The ratio of the $e.m.f.$s of the cells is:

$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire is (in $m$)

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