A resistance of 2 Omega is connected across o | vedclass

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

Question

(A) Let the unknown resistance be $x$. According to the meter bridge principle,for the first case:$\frac{2}{x} = \frac{\ell}{100 - \ell}$ ............

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let the unknown resistance be $x$. According to the meter bridge principle,for the first case:$\frac{2}{x} = \frac{\ell}{100 - \ell}$ ...............$(I)$When the resistances are interchanged,the new balance point is $\ell' = \ell + 20$ (since $x > 2$,the balance point shifts towards the larger resistance). Thus,for the second case:$\frac{x}{2} = \frac{\ell + 20}{100 - (\ell + 20)} = \frac{\ell + 20}{80 - \ell}$ ...............$(II)$From $(I)$,$\frac{\ell}{100 - \ell} = \frac{2}{x} \implies \ell x = 200 - 2\ell \implies \ell(x + 2) = 200 \implies \ell = \frac{200}{x + 2}$.From $(II)$,$\frac{x}{2} = \frac{\ell + 20}{80 - \ell} \implies 80x - \ell x = 2\ell + 40 \implies 80x - 40 = \ell(x + 2)$.Substituting $\ell(x + 2) = 200$ into the equation $80x - 40 = \ell(x + 2)$:$80x - 40 = 200$$80x = 240$$x = 3 \Omega$.

Similar Questions

During an experiment with a meter bridge,the galvanometer shows a null point when the jockey is pressed at $40.0 \ cm$ using a standard resistance of $90 \ \Omega$,as shown in the figure. The least count of the scale used in the meter bridge is $1 \ mm$. The unknown resistance is:

In a meter bridge experiment,initially the jockey is at the null point. Now,resistances $R_1$ and $R_2$ are interchanged. The shift in the position of the jockey is ................ $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module