(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 →

If the null point of a meter bridge is obtained at $50 \ cm$,what will be the value of the unknown resistor $X$ if the known resistor $R$ is $10 \ \Omega$ (in $\Omega$)?

Medium

View Solution

In a meter bridge experiment,the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $X \Omega$ is added to the $3 \Omega$ resistor to shift the null point by $22.5 \text{ cm}$. The value of $X$ is: (in $\Omega$)

MediumMHT CET 2024

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 $\Omega$)

DifficultAP EAMCET 2013

View Solution

In the experimental setup of a meter bridge shown in the figure,the null point is obtained at a distance of $40\,cm$ from $A$. If a $10\,\Omega$ resistor is connected in series with $R_1$,the null point shifts by $10\,cm$. The resistance that should be connected in parallel with $(R_1 + 10)\,\Omega$ such that the null point shifts back to its initial position is .............. $\Omega$.

DifficultJEE MAIN 2019

View Solution

Two wires $A$ and $B$ of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at $40 \text{ cm}$ from left end. Diameters of the wires $A$ and $B$ are in the ratio $3:1$ respectively, the ratio of specific resistance of $A$ to that of $B$ is (in $: 1$)

DifficultMHT CET 2026

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

If the null point of a meter bridge is obtained at $50 \ cm$,what will be the value of the unknown resistor $X$ if the known resistor $R$ is $10 \ \Omega$ (in $\Omega$)?

In a meter bridge experiment,the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $X \Omega$ is added to the $3 \Omega$ resistor to shift the null point by $22.5 \text{ cm}$. The value of $X$ is: (in $\Omega$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module