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Question

(B) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L} = \frac{I \cdot R}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \

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$4$

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(B) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L} = \frac{I \cdot R}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \frac{I \cdot \rho}{A}$,where $I$ is the current,$\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.However,in a standard potentiometer circuit,the potential drop across the wire is $V_{wire} = I \cdot R_{wire} = I \cdot \frac{\rho \cdot L}{A}$.When the cell is balanced,the emf of the cell $E = k \cdot l$,where $l$ is the balancing length.If the length $L$ of the potentiometer wire is kept constant and the cross-section $A$ is doubled,the resistance of the wire $R = \rho \frac{L}{A}$ becomes half.Since the battery voltage is steady,the current $I$ through the wire increases,but the potential drop across the wire remains equal to the battery voltage.Thus,the potential gradient $k = \frac{V_{battery}}{L}$ remains unchanged because both $V_{battery}$ and $L$ are constant.Since $E = k \cdot l$ and both $E$ and $k$ are constant,the balancing length $l$ remains $4 \,m$.

$A$ $10 \,m$ long potentiometer wire is connected to a battery having a steady voltage. $A$ Leclanche cell is balanced at $4 \,m$ length of the wire. If the length is kept the same,but its cross-section is doubled,the null point will be obtained at ........... $m$.

Similar Questions

In a potentiometer,a balance point is obtained when:

$A$ potentiometer wire is $4 \,m$ long and a potential difference of $3 \,V$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \,cm$ of the potentiometer wire is: (in $\,V$)

$A$ $10 \, m$ long wire of $20 \, \Omega$ resistance is connected with a battery of $3 \, V$ $e.m.f.$ (negligible internal resistance) and a $10 \, \Omega$ resistance is joined to it in series. The potential gradient along the wire in $V/m$ is:

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 the given potentiometer circuit,the balancing length for points $B$ and $C$ is $40 \, cm$. What is the balancing length for points $C$ and $D$ in $cm$?

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