(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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The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

Medium

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In a potentiometer experiment,cells of e.m.f. $E_{1}$ and $E_{2}$ are connected in series $(E_{1} > E_{2})$,and the balancing length is $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

MediumMHT CET 2020

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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:

Medium

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In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (in $V$)

Medium

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$A$ potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. $A$ cell $C$ having $emf$ $\varepsilon/2$ and internal resistance $3r$ is connected as shown in the figure. The length $AJ$ at which the galvanometer shows no deflection is

MediumJEE MAIN 2019

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The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

In a potentiometer experiment,cells of e.m.f. $E_{1}$ and $E_{2}$ are connected in series $(E_{1} > E_{2})$,and the balancing length is $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

$A$ 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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