(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

Medium

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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$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)

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In a potentiometer circuit,a cell of $2\,V$ $e.m.f.$ and $5\,\Omega$ internal resistance is connected to a uniform wire of length $1000\,cm$ and resistance $15\,\Omega$. The potential gradient of the wire is:

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In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.

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For a cell of $e.m.f.$ $2\,V$,a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega$ resistor and the balance is obtained across $40\, cm$ of the wire,then the internal resistance of the cell is ............. $\Omega$.

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In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

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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

$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)

In a potentiometer circuit,a cell of $2\,V$ $e.m.f.$ and $5\,\Omega$ internal resistance is connected to a uniform wire of length $1000\,cm$ and resistance $15\,\Omega$. The potential gradient of the wire is:

In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.

For a cell of $e.m.f.$ $2\,V$,a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega$ resistor and the balance is obtained across $40\, cm$ of the wire,then the internal resistance of the cell is ............. $\Omega$.

In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

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