$A$ galvanometer has a resistance of $100 \Omega$ and a current of $10 \text{ mA}$ produces full-scale deflection in it. The resistance to be connected to it in series,to convert it into a voltmeter of range $50 \text{ V}$,is: (in $Omega$)

Explain the solution to the difficulties that arise when using a galvanometer directly as an ammeter.

$A$ 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 $\Omega$)

The current flowing through a coil of resistance $900 \, \Omega$ is to be reduced by $90 \, \%$. What value of shunt should be connected across the coil in $\Omega$?

The deflection in a moving coil galvanometer of resistance $45 \Omega$ falls from $30$ divisions to $3$ divisions. The length of the shunt wire required to convert the galvanometer into an ammeter is [specific resistance of the material of the shunt wire $= 5 \times 10^{-7} \Omega m$ and area of cross-section of the wire $= 4 \times 10^{-7} m^2$]. (in $m$)

An ammeter with a resistance of $1\, \Omega$ can measure up to $10\, mA$. To convert it into a voltmeter that can measure up to $10\, V$,what resistance (in $\Omega$) must be connected in series?