$A$ galvanometer has a resistance of $50 \ \Omega$ and it allows a maximum current of $5 \ mA$. It can be converted into a voltmeter to measure up to $100 \ V$ by connecting in series a resistor of resistance: (in $Omega$)

An ammeter with a resistance of $2 \,\Omega$ can measure up to $100 \,mA$. What resistance (in $\Omega$) must be connected in series to make it measure up to $5 \,V$?

The galvanometer has a resistance of $1.8 \Omega$. Calculate the value of shunt to increase the range of galvanometer by $10$ times. (in $Omega$)

To find the resistance of a galvanometer by the half-deflection method,the experimental data obtained are given in the table below:

$S. No.$	Resistance $R \ (\Omega)$	Deflection $(\theta)$	Shunt $S \ (\Omega)$	Half deflection $(\theta / 2)$	Galvanometer resistance $(G)$
$1$	$3300$	$30$	$80$	$15$	$G_1$
$2$	$5000$	$20$	$80$	$10$	$G_2$

From the above data,the galvanometer resistance $G$ will be near to: (in $\Omega$)

Why is the total resistance of a voltmeter kept very high?

If only $\frac{1}{51}$ of the main current is to be passed through a galvanometer,then the shunt required is $R_1$. If only $\frac{1}{11}$ of the main voltage is to be developed across the galvanometer,then the resistance required is $R_2$. Find the ratio $\frac{R_2}{R_1}$.