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 $5 \,V$?

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series,showing a full-scale deflection of $30$ divisions. The additional series resistance required to reduce the deflection to $20$ divisions is: (in $\Omega$)

$A$ galvanometer of resistance $G \ \Omega$ is shunted by a resistance $S \ \Omega$. To keep the main current in the circuit unchanged,the resistance to be connected in series with the galvanometer is

In the circuit shown below,the ammeter and the voltmeter readings are $3 \text{ A}$ and $6 \text{ V}$ respectively. Then the value of the resistance $R$ is

$A$ voltmeter of $250 mV$ range having a resistance of $10 \Omega$ is converted into an ammeter of $250 mA$ range. The value of the necessary shunt is (nearly): (in $\Omega$)

With a resistance $X$ connected in series with a galvanometer of resistance $100 \Omega$,it acts as a voltmeter of range $0-15 \ V$. To double the range,a resistance of $1500 \Omega$ is to be connected in series with $X$. The value of $X$ in ohm is: