Q 12-03-273JEE MainJEE Main 2019 (12 Jan, Shift 1)Easy
An ideal battery of emf $4$ V and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1$ m and resistance $5\ \Omega$. The value of $R$, to give a potential difference of $5$ mV across $10$ cm of potentiometer wire, is:
Answer: (C) $395\ \Omega$
$5$ mV across $10$ cm means $50$ mV across the whole $1$ m wire, so the current is $\dfrac{0.05}{5} = 0.01$ A.
$$\frac{4}{5 + R} = 0.01 \Rightarrow R = 395\ \Omega$$
Solution by Sreeraj P, M.Sc Physics