To send $10 \%$ of the main current through a moving coil galvanometer of resistance $99 \Omega$,the shunt required is . . . . . . . (in $\Omega$)

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter reading up to $10\,A$,it is necessary to put a resistance of

$A$ certain current passing through a galvanometer produces a deflection of $100$ divisions. When a shunt of $1 \ \Omega$ is connected, the deflection reduces to $1$ division. The galvanometer resistance is: (in $\Omega$)

In an experiment to find the resistance of a galvanometer by the half-deflection method, a $5 V$ battery and a high resistance of $4.9 k\Omega$ are connected in the circuit. In the absence of any shunt resistance, the galvanometer reads $20$ divisions when current flows in the circuit. To reduce the deflection by half, the value of the shunt resistance used is $98 \Omega$. The figure of merit of the galvanometer is given as $..... \mu A / \text{division}$.

The voltmeter has a range of $10 \ V$ and its internal resistance is $50 \ \Omega$. To increase the range of the voltmeter to $15 \ V$,the resistance which is to be connected is:

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is