When the two known resistances $R$ and $S$ are connected in the left and right gaps of a meter bridge respectively,the null point is found at a distance $l_1$ from the zero end of the meter bridge wire. An unknown resistance $X$ is now connected in parallel with $S$,and the null point is found at a distance $l_2$ from the zero end of the meter bridge wire. The unknown resistance $X$ is:
- A$\frac{S l_1(100-l_2)}{100(l_2-l_1)}$
- B$\frac{S l_2(100-l_1)}{100(l_1-l_2)}$
- C$\frac{100(l_2-l_1)}{S l_1(100-l_2)}$
- D$\frac{100(l_2-l_1)}{S l_2(100-l_1)}$
Explore More
Similar Questions
In a meter bridge,the balancing length from the left end (standard resistance of $1 \, \Omega$ is in the right gap) is found to be $20 \, cm$. The value of the unknown resistance is ............... $\Omega$.
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$A$ student obtained the following observations in an experiment of a meter bridge to find the unknown resistance of the circuit. The most accurate value of the unknown resistance is ............ $\Omega$.
$S.No.$ $R$ $l$ $100-l$ $S = \left( \frac{100-l}{l} \right)R$ $1$ $20\,\Omega$ $43$ $57$ $26.51\,\Omega$ $2$ $30\,\Omega$ $51$ $49$ $28.82\,\Omega$ $3$ $40\,\Omega$ $59$ $41$ $27.80\,\Omega$ $4$ $60\,\Omega$ $70$ $30$ $25.71\,\Omega$
| $S.No.$ | $R$ | $l$ | $100-l$ | $S = \left( \frac{100-l}{l} \right)R$ |
| $1$ | $20\,\Omega$ | $43$ | $57$ | $26.51\,\Omega$ |
| $2$ | $30\,\Omega$ | $51$ | $49$ | $28.82\,\Omega$ |
| $3$ | $40\,\Omega$ | $59$ | $41$ | $27.80\,\Omega$ |
| $4$ | $60\,\Omega$ | $70$ | $30$ | $25.71\,\Omega$ |
Medium
View SolutionIn a meter-bridge,if the left and right gaps are connected with $2 \Omega$ and $3 \Omega$ resistances,respectively,then the bridge is balanced. What resistance should be connected with the $3 \Omega$ resistance to get the balancing point at the midpoint of the bridge wire?
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 SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo