Q 12-03-229JEE MainJEE Main 2020 (8 Jan, Shift 1)Medium
The length of a potentiometer wire is $1200$ cm and it carries a current of $60$ mA. For a cell of emf $5$ V and internal resistance of $20\ \Omega$, the null point on it is found to be at $1000$ cm. The resistance of the whole wire is:
Answer: (D) $100\ \Omega$
At the null point no current flows through the cell, so its internal resistance plays no part and the potential drop along $1000$ cm equals the emf:
$$\text{potential gradient} = \frac{5\ \text{V}}{1000\ \text{cm}}$$
Potential drop across the whole wire: $\dfrac{5}{1000}\times1200 = 6$ V.
$$R = \frac{6}{60\times10^{-3}} = 100\ \Omega$$
(The official answer key lists $120\ \Omega$.)
Solution by Sreeraj P, M.Sc Physics