(B) For a potentiometer,the $emf$ of a cell is directly proportional to the balancing length $(l)$: $E \propto l$,which implies $\frac{E_1}{E_2} = \fr

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(B) For a potentiometer,the $emf$ of a cell is directly proportional to the balancing length $(l)$: $E \propto l$,which implies $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.Given $E_1 = 1.5 \ V$ and $l_1 = 60 \ cm$.The new length $l_2 = l_1 + 40 \ cm = 60 + 40 = 100 \ cm$.Substituting the values: $\frac{1.5}{E} = \frac{60}{100}$.$\frac{1.5}{E} = \frac{6}{10} = 0.6$.$E = \frac{1.5}{0.6} = \frac{15}{6} = 2.5 \ V$.Given $E = \frac{x}{10} \ V$,so $\frac{x}{10} = 2.5$.$x = 25$.

Class 12 Physics Current Electricity Potentiometer

Medium

In an experiment to find the $emf$ of a cell using a potentiometer,the length of the null point for a cell of $emf$ $1.5 \ V$ is found to be $60 \ cm$. If this cell is replaced by another cell of $emf$ $E$,the length of the null point increases by $40 \ cm$. The value of $E$ is $\frac{x}{10} \ V$. The value of $x$ is $............$

JEE MAIN 2023 All JEE MAIN

A
$24$
B
$25$
C
$23$
D
$22$

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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:

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

DifficultMHT CET 2026

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In the experiment of a potentiometer,at balance,there is no current in the

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The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is

DifficultTS EAMCET 2002

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$A$ potentiometer wire,$10 \, m$ long,has a resistance of $40 \, \Omega$. It is connected in series with a resistance box and a $2 \, V$ storage cell. If the potential gradient along the wire is $0.1 \, mV/cm$,the resistance unplugged in the box is .............. $\Omega$.

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

In the experiment of a potentiometer,at balance,there is no current in the

The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is

$A$ potentiometer wire,$10 \, m$ long,has a resistance of $40 \, \Omega$. It is connected in series with a resistance box and a $2 \, V$ storage cell. If the potential gradient along the wire is $0.1 \, mV/cm$,the resistance unplugged in the box is .............. $\Omega$.

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