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When a conducting wire is connected in the left gap and a known resistance in the right gap, of a metre bridge the balancing length is 80 cm. If the wire is cut into 2 equal parts and only one part is connected in the left gap, without changing the resistance in the other gap then the balancing length

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Shifts right by 13.3 cm

Shifts left by 45.7 cm

Shifts right by 45.7 cm

Shifts left by 13.3 cm

answer is A.

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Detailed Solution

$\frac{X}{R} = \frac{80}{20} \dots \dots (1)$

$\frac{X}{2 R} = \frac{l}{100 - l} \Rightarrow 200 - 2 l = l$

$l = 66.67 cm$

$\Rightarrow Δ l = (80 - 66.7) cm$

$= 13.3 cm to l eft$

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