The length of a potentiometer wire is 1200   c m and it carries a current of 60   m A For a cell…

1. $80\mathrm{\Omega }$
2. $120\mathrm{\Omega }$
3. $60\mathrm{\Omega }$
4. $100\mathrm{\Omega }$

Solution

Potential gradient $=\frac{5}{1000}=\frac{V_P}{1200}$
$V_P=6 V$
And $R_P=\frac{V_P}{I}=\frac{6}{60\times {10}^{-3}}=100 \Omega$

Asked in: JEE Main 2020 (08 Jan Shift 1)

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