$A$ galvanometer having a coil resistance of $100 \ \Omega$ gives a full-scale deflection when a current of $1 \ mA$ is passed through it. The value of the shunt resistance,which can convert this galvanometer into an ammeter giving a full-scale deflection for a current of $10 \ A$,is: ...... $\Omega$

The magnets of two suspended coil galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side $a$ and that of the second one is circular of radius $\frac{a}{\sqrt{\pi}}.$ When the same current is passed through the coils,the ratio of the torque experienced by the first coil to that experienced by the second one is

$A$ galvanometer of resistance $G$ is shunted by a resistance of $5 \Omega$. To keep the main current in the circuit unchanged,the resistance to be put in series with the same galvanometer is

$A$ galvanometer coil has a resistance of $12\; \Omega$ and the meter shows full-scale deflection for a current of $3\; mA$. How will you convert the meter into a voltmeter of range $0$ to $18\; V$?

$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.

$A$ galvanometer has a resistance of $80 \Omega$ and it is shunted with a resistance of $20 \Omega$. If $20 \%$ of the main current flows through the galvanometer, what is the value of the main current (in $\text{ A}$)?