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(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

$3$

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(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$.

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