A galvanometer of $50 \Omega$ resistance is converted into an ammeter using a shunt resistance of $10 \Omega$. If the same resistance is used to convert the same galvanometer in to a voltmeter, what would be the ratio of the resistance of the ammeter to the resistance of the voltmeter?

Let $G = 50 \Omega$ be the resistance of the galvanometer and $S = 10 \Omega$ be the shunt resistance.

The resistance of an ammeter $R_{A}$ is given by the parallel combination of the galvanometer and the shunt resistance:

$R_{A} = \dfrac{G \times S}{G + S} = \dfrac{50 \times 10}{50 + 10} = \dfrac{500}{60} = \dfrac{50}{6} \Omega = \dfrac{25}{3} \Omega$.

When the same resistance $S = 10 \Omega$ is used to convert the galvanometer into a voltmeter, it is connected in series with the galvanometer. The resistance of the voltmeter $R_{V}$ is given by:

$R_{V} = G + S = 50 + 10 = 60 \Omega$.

The ratio of the resistance of the ammeter to the resistance of the voltmeter is:

$\dfrac{R_{A}}{R_{V}} = \dfrac{25/3}{60} = \dfrac{25}{3 \times 60} = \dfrac{25}{180}$.

Dividing both numerator and denominator by 5, we get:

$\dfrac{R_{A}}{R_{V}} = \dfrac{5}{36}$.

Answer: $5: 36$