Q 12-04-065JEE MainJEE Main 2024 (1 Feb, Shift 1)Easy
A galvanometer has a resistance of $50\ \Omega$ and allows a maximum current of $5\ \text{mA}$. It can be converted into a voltmeter to measure up to $100\ \text{V}$ by connecting in series a resistor of resistance
Answer: (C) $19950\ \Omega$
$$R = \frac{V}{I_g} - G = \frac{100}{5\times10^{-3}} - 50 = 20000 - 50 = 19950\ \Omega$$
Solution by Sreeraj P, M.Sc Physics