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The length of a potentiometer wire is 1200cm and it carries a current of 60mA. For a cell of emf 5V and internal resistance of 20 $Ω$ , the null point on it is found to be at 1000cm. the resistance of whole wire is:

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80 $Ω$

120 $Ω$

60 $Ω$

100 $Ω$

answer is D.

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Detailed Solution

Null point at 1000 cm means , potential across 1000 cm is 5V.

Let say potential across 1200 cm is $V_{p}$

Thus, potential gradient = $\frac{5}{1000} = \frac{V_{p}}{1200}$ $\Rightarrow V_{P} = 6 V$

Now, use Ohm's Law across the whole wire,

$R_{P} = \frac{V_{P}}{I} = \frac{6}{60 \times 10^{- 3}} = 100 Ω$

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