The value of shunt resistance,that allows only $10 \%$ of the main current to pass through a galvanometer of resistance $99 \Omega$,is (in $\Omega$)

If an ammeter is to be used in place of a galvanometer,then we must connect:

$A$ voltmeter has a range $V$ and resistance $G \, \Omega$. The value of the series resistance required to convert this voltmeter into a voltmeter of range $nV$ is ....

$A$ galvanometer coil has a resistance $90\, \Omega$ and full-scale deflection current $10\, mA$. $A$ $910\, \Omega$ resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.1\, V$,the number of divisions on its scale is:

The coil of a galvanometer consists of $100$ turns and an effective area of $1 \, cm^2$. The restoring couple per unit radian is $10^{-8} \, N \cdot m/rad$. The magnetic field between the pole pieces is $5 \, T$. The current sensitivity of this galvanometer will be:

$A$ milliammeter of resistance $40 \Omega$ has a range $0-30 \text{ mA}$. What will be the resistance used in series to convert it into a voltmeter of range $0-15 \text{ V}$ (in $\Omega$)?