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:

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

An ideal ammeter and an ideal voltmeter have resistances of . . . . . . $\Omega$ and . . . . . . $\Omega$ respectively.

Two galvanometers $A$ and $B$ require $3\,mA$ and $5\,mA$ respectively to produce the same deflection of $10$ divisions. Then:

When a current of $5 \ mA$ is passed through a galvanometer having a coil of resistance $15 \ \Omega$,it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range $0 - 10 \ V$ is

If only $1 \%$ of the total current is passed through a galvanometer of resistance $G$,then the resistance of the shunt is: