(A) In a potentiometer,the balancing length $\ell$ is directly proportional to the emf $\varepsilon$ of the cell,given by $\varepsilon \propto \ell$ o

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) In a potentiometer,the balancing length $\ell$ is directly proportional to the emf $\varepsilon$ of the cell,given by $\varepsilon \propto \ell$ or $\frac{\varepsilon_1}{\varepsilon_2} = \frac{\ell_1}{\ell_2}$.Given $\varepsilon_1 : \varepsilon_2 = 3 : 2$ and $\ell_1 = 75 \, cm$.Substituting the values: $\frac{3}{2} = \frac{75}{\ell_2}$.Solving for $\ell_2$: $\ell_2 = \frac{75 \times 2}{3} = 50 \, cm$.The difference in the balancing lengths is $\Delta \ell = |\ell_1 - \ell_2| = |75 - 50| = 25 \, cm$.

Class 12 Physics Current Electricity Potentiometer

Medium

In a potentiometer arrangement,a cell gives a balancing point at $75 \, cm$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $3:2$,the difference in the balancing length of the potentiometer wire in above two cases will be ......... $cm$.

JEE MAIN 2022 All JEE MAIN

A
$25$
B
$255$
C
$95$
D
$27$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Current Electricity

Subtopic

Potentiometer

Previous Year

JEE MAIN 2022 Paper All JEE MAIN years →

$A$ cell in a 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 $m$)

DifficultTS EAMCET 2009

View Solution

$A$ potentiometer is an ideal device for measuring potential difference because

Easy

View Solution

The length of a potentiometer wire is $l$. $A$ cell of $emf$ $E$ is balanced at a length $l/3$ from the positive end of the wire. If the length of the wire is increased by $l/2$,at what distance will the same cell give a balance point?

Difficult

View Solution

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

View Solution

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is:

MediumNEET 2016

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ cell in a 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 $m$)

$A$ potentiometer is an ideal device for measuring potential difference because

The length of a potentiometer wire is $l$. $A$ cell of $emf$ $E$ is balanced at a length $l/3$ from the positive end of the wire. If the length of the wire is increased by $l/2$,at what distance will the same cell give a balance point?

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module