(A) In a meter bridge,the balanced condition is given by $\frac{P}{Q} = \frac{l_1 + \alpha}{l_2 + \beta}$,where $P = 15\,\Omega$,$Q = R$,$l_1 = 43\,cm

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

(A) In a meter bridge,the balanced condition is given by $\frac{P}{Q} = \frac{l_1 + \alpha}{l_2 + \beta}$,where $P = 15\,\Omega$,$Q = R$,$l_1 = 43\,cm$,$l_2 = (100 - 43) = 57\,cm$,$\alpha = 2\,cm$ (end correction at $A$),and $\beta = 1\,cm$ (end correction at $B$).Substituting the values:$\frac{15}{R} = \frac{43 + 2}{57 + 1}$$\frac{15}{R} = \frac{45}{58}$$R = \frac{15 \times 58}{45}$$R = \frac{58}{3} \approx 19.33\,\Omega$.Given the options,the closest integer value is $19\,\Omega$.

Class 12 Physics Current Electricity Meter Bridge

Difficult

$A$ meter bridge setup is shown in the figure. It is used to determine an unknown resistance $R$ using a given resistor of $15\,\Omega$. The galvanometer $(G)$ shows null deflection when the tapping key is at the $43\,cm$ mark from end $A$. If the end correction for end $A$ is $2\,cm$ and for end $B$ is $1\,cm$,then the determined value of $R$ will be . . . . . . $\Omega$.

JEE MAIN 2022 All JEE MAIN

A
$19$
B
$20$
C
$24$
D
$18$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Current Electricity

Subtopic

Meter Bridge

Previous Year

JEE MAIN 2022 Paper All JEE MAIN years →

The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The smaller resistance is (in $\Omega$)

MediumMHT CET 2020

View Solution

When the right gap of a meter bridge consists of two equal resistors in series,the balancing point is at $50 \ cm$. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap,the balancing point is at: (in $cm$)

MediumTS EAMCET 2025

View Solution

In the meter bridge experiment,the length $AB$ of the wire is $1 \ m$. The resistors $X$ and $Y$ have values $5 \ \Omega$ and $2 \ \Omega$ respectively. When a shunt resistance $S$ is connected in parallel to $X$,the balancing point is found to be $0.625 \ m$ from $A$. Then,the resistance of the shunt $S$ is (in $\Omega$)

DifficultTS EAMCET 2013

View Solution

In a meter bridge circuit, which part is called the battery arm and which part is called the galvanometer arm?

Medium

View Solution

In a meter bridge experiment,a null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then where will be the new position of the null point from the same end,if one decides to balance a resistance of $4X$ against $Y$ (in $cm$)?

MediumAIEEE 2004

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

The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The smaller resistance is (in $\Omega$)

When the right gap of a meter bridge consists of two equal resistors in series,the balancing point is at $50 \ cm$. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap,the balancing point is at: (in $cm$)

In a meter bridge circuit, which part is called the battery arm and which part is called the galvanometer arm?

Vedclass Test Series

Exam Paper Generator

Online Exam Module