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Q.

In the shown arrangement of a meter bridge, if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled?

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a

x

b

2x

c

x/4

d

4x

answer is A.

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Detailed Solution

$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}} = \frac{x}{100 - x}$ If the radius of the wire is doubled then the resistance AC is changed so does the value of resistance CB. So $\frac{R_{1}}{R_{2}}$ remains unchanged.

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