MHTCET 16 May 2024 Shift 1 Paper

Concept:The voltage range of a galvanometer is given by $V = I_g (G + R_s)$, where $I_g$ is the full-scale deflection current and $R_s$ is the series resistance.Explanation:For the first case, a series resistance of $100\ \Omega$ gives range $V$, so:$$V = I_g (G + 100)$$For the second case, the range is doubled to $2V$ using a series resistance of $1000\ \Omega$, so:$$2V = I_g (G + 1000)$$Subtract the first equation from the second:$$2V - V = I_g[(G + 1000) - (G + 100)]$$$$V = I_g \times 900$$Substitute this value of $V$ into the first equation:$$I_g \times 900 = I_g (G + 100)$$Cancelling $I_g$ from both sides:$$900 = G + 100$$Solving for $G$:$$G = 800\ \Omega$$Answer:The galvanometer resistance is $800\ \Omega$. Hence, option A is correct.