$A$ voltmeter having a resistance of $50 \times 10^3 \, \Omega$ is used to measure the voltage in a circuit. To increase the range of measurement $3$ times,the additional series resistance required is:

$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

$A$ current of $10^{-3} \ A$ is flowing through a resistance of $1000 \ \Omega$. To measure the correct potential difference across this resistance,a voltmeter must be used whose resistance should be:

$A$ galvanometer has $30$ divisions and a current sensitivity of $20 \, \mu A$ per division. Its resistance is $25 \, \Omega$. How will you convert this galvanometer into a voltmeter of $1 \, V$ range? (Find the series resistance in $\Omega$)

The number of turns in a coil of a galvanometer is tripled. Then:

$A$ 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: ...... $\Omega$