In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

In an experiment to find the resistance of a galvanometer by the half-deflection method, a $5 V$ battery and a high resistance of $4.9 k\Omega$ are connected in the circuit. In the absence of any shunt resistance, the galvanometer reads $20$ divisions when current flows in the circuit. To reduce the deflection by half, the value of the shunt resistance used is $98 \Omega$. The figure of merit of the galvanometer is given as $..... \mu A / \text{division}$.

An ammeter of resistance $R$ gives a full-scale deflection when a current of $2 \ A$ passes through it. If it is to be converted into an ammeter to measure a maximum current of $10 \ A$,the required shunt is:

To make the magnetic field radial in a moving coil galvanometer:

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 $\Omega$)

$A$ galvanometer has a current range of $10 \text{ mA}$ and a voltage range of $0.75 \text{ V}$. To convert this galvanometer into an ammeter of range $10 \text{ A}$, what is the shunt resistance?