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Q.

In given figure, the potentiometer wire AB has a resistance of $5 Ω$ and length 10 m. The balancing length AM for the emf of 0.4 V is

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a

4 m

b

0.8 m

c

8 m

d

0.4 m

answer is D.

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Detailed Solution

Emf of cell, $E = Xl = \frac{V}{l} = \frac{iR}{L} \times l$

$∴ E = \frac{e}{(R + r)} \times \frac{r}{L} \times l$

here R=45 $Ω$ , r=5 $Ω$ , L=10m, l= balancing distance

on solving we get

$∴ l = 8 m$

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