(A) In a potentiometer, the potential drop per unit length $(\phi)$ is constant.The balance length $(l)$ is directly proportional to the $EMF$ $(E)$ o

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(A) In a potentiometer, the potential drop per unit length $(\phi)$ is constant.The balance length $(l)$ is directly proportional to the $EMF$ $(E)$ of the cell, i.e., $E = \phi l$.Therefore, the ratio is given by: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.Given: $E_1 = 1.5\, V$, $l_1 = 36\, cm$, $E_2 = 2.5\, V$.Substituting the values: $\frac{1.5}{2.5} = \frac{36}{l_2}$.Simplifying the ratio: $\frac{3}{5} = \frac{36}{l_2}$.Solving for $l_2$: $l_2 = \frac{36 \times 5}{3} = 12 \times 5 = 60\, cm$.

Class 12 Physics Current Electricity Potentiometer

Medium

In a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)

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$60$
B
$21.6$
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$64$
D
$62$

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$A$ potentiometer has a wire of length $4 \ m$ and resistance $10 \ \Omega$. The potentiometer is connected to a cell of $2 \ V$. The potential drop per unit length is ........... $V/m$.

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For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because:

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$A$ $2\,V$ battery,a $15\,\Omega$ resistor,and a potentiometer of $100\,cm$ length are all connected in series. If the resistance of the potentiometer wire is $5\,\Omega$,then the potential gradient of the potentiometer wire is ............... $V/cm$.

MediumAIIMS 1982

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The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.

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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 a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)

Similar Questions

$A$ potentiometer has a wire of length $4 \ m$ and resistance $10 \ \Omega$. The potentiometer is connected to a cell of $2 \ V$. The potential drop per unit length is ........... $V/m$.

For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because:

$A$ $2\,V$ battery,a $15\,\Omega$ resistor,and a potentiometer of $100\,cm$ length are all connected in series. If the resistance of the potentiometer wire is $5\,\Omega$,then the potential gradient of the potentiometer wire is ............... $V/cm$.

The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.

$A$ 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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