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

Option 1 - <math> <mn>60</mn> <mi mathvariant="normal">Ω</mi> </math>

Option 2 - <math> <mn>100</mn> <mi mathvariant="normal">Ω</mi> </math>

Option 3 - <math> <mn>80</mn> <mi mathvariant="normal">Ω</mi> </math>

Option 4 - <math> <mn>120</mn> <mi mathvariant="normal">Ω</mi> </math>

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Raj Pandey | Contributor-Level 9

6 months ago

Correct Option - 2

Detailed Solution:

Potential gradient $= \frac{5}{1000} = \frac{V_{P}}{1200}$

$V_{P} = 6 V$ $\frac{}{} \frac{_{}}{}$

and $R_{P} = \frac{V_{P}}{l} = \frac{6}{60 * 10^{- 3}} = 100 Ω$ $_{} \frac{_{}}{} \frac{}{^{}}$

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

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