A galvanometer of resistance \( G \) is connected in series with a resistor of high resistance \( R \) such that the potential drop across the galvanometer is \( (1/10) \)th of the total applied voltage. The combined resistance of the galvanometer and the series resistor is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 10\ G \)
In series the same current \( I \) flows through the galvanometer and the resistor, so the voltage divides in the ratio of the resistances.
\( V_{G} = IG = \frac{V}{10} \)
\( \displaystyle R_{\mathrm{total}} = \frac{V}{I} = \frac{10\,IG}{I} = 10G \)
So \( R = 9G \), and the combination (galvanometer + series resistor, that is, a voltmeter) has resistance \( 10G \).