When unknown resistance and a resistance of 4 $Ω$ are used in the left and right gaps of a meter bridge, the balance point is 50cm. The shift in the balance point if a 4 $Ω$ resistance is now connected parallel to the resistor in right gap

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$\frac{100}{3} cm$

$\frac{50}{3} cm$

$\frac{300}{3} cm$

$\frac{400}{5} cm$

answer is B.

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Detailed Solution

$\frac{x}{4} = \frac{50}{100 - 50}, ∴ x = 4 Ω$
Question Image
$\frac{4}{2} = \frac{ℓ^{1}}{100 - ℓ^{1}} solve ℓ^{1} = \frac{200}{3} cm$
Shift $= \frac{200}{3} - 50 = \frac{50}{3} cm$

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