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Q.

In a meter bridge, the null point is obtained at 40 cm division. When an identical resistance is connected in parallel the smaller one, the balance point shifts by

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a

20 cm

b

25 cm

c

15 cm

d

10 cm

answer is A.

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Detailed Solution

$\frac{X}{R} = \frac{40}{60} = \frac{2}{3}$

$\frac{(\frac{X}{2})}{R} = (\frac{l}{100 - l}) \Rightarrow \frac{1}{3} = (\frac{l}{100 - l})$

$l = 25 c m \Rightarrow Δ l = 15 c m$

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