(B) Let $X'$ be the equivalent resistance of $X$ and $S$ connected in parallel. $X' = \frac{X \cdot S}{X + S} = \frac{5S}{5 + S}$. In a meter bridge,t

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

(B) Let $X'$ be the equivalent resistance of $X$ and $S$ connected in parallel. $X' = \frac{X \cdot S}{X + S} = \frac{5S}{5 + S}$. In a meter bridge,the balancing condition is given by $\frac{X'}{Y} = \frac{l_1}{l_2}$,where $l_1 = 0.625 \ m$ and $l_2 = 1 - 0.625 = 0.375 \ m$. Substituting the values: $\frac{5S / (5 + S)}{2} = \frac{0.625}{0.375}$ $\frac{5S}{2(5 + S)} = \frac{625}{375} = \frac{5}{3}$ $15S = 10(5 + S)$ $15S = 50 + 10S$ $5S = 50$ $S = 10 \ \Omega$.

Class 12 Physics Current Electricity Meter Bridge

Difficult

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 $Omega$)

AP EAMCET 2013 All AP EAMCET

TS EAMCET 2013 All TS EAMCET

A
$5$
B
$10$
C
$7.5$
D
$12.5$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Current Electricity

Subtopic

Meter Bridge

Previous Year

AP EAMCET 2013 Paper All AP EAMCET years →

Previous Year

TS EAMCET 2013 Paper All TS EAMCET years →

Explain how the value of an unknown resistor can be obtained by using a meter bridge.

Difficult

View Solution

$A$ meter bridge with two resistances $R_{1}$ and $R_{2}$ as shown in the figure was balanced (null point) at $40 \text{ cm}$ from the point $P$. The null point changed to $50 \text{ cm}$ from the point $P$,when a $16 \ \Omega$ resistance is connected in parallel to $R_{2}$. The values of resistances $R_{1}$ and $R_{2}$ are . . . . . . .

DifficultJEE MAIN 2026

View Solution

In a meter bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)

DifficultTS EAMCET 2007

View Solution

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40\,cm$. Now an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $80\,cm$ measured from the same end. Then the value of $x$ will be $.......\Omega$.

DifficultJEE MAIN 2022

View Solution

In a meter bridge experiment,when a nichrome wire is in the right gap,the balancing length is $60 \ cm$. When the nichrome wire is uniformly stretched to increase its length by $20 \%$,and again connected in the right gap,the new balancing length is nearly: (in $cm$)

MediumAP EAMCET 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

Explain how the value of an unknown resistor can be obtained by using a meter bridge.

In a meter bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)

In a meter bridge experiment,when a nichrome wire is in the right gap,the balancing length is $60 \ cm$. When the nichrome wire is uniformly stretched to increase its length by $20 \%$,and again connected in the right gap,the new balancing length is nearly: (in $cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module