$A$ voltmeter of $250 mV$ range having a resistance of $10 \Omega$ is converted into an ammeter of $250 mA$ range. The value of the necessary shunt is (nearly): (in $\Omega$)

In an ammeter,$4 \%$ of the main current is passing through the galvanometer. If the shunt resistance is $5 \Omega$,then the resistance of the galvanometer will be: (in $\Omega$)

An ammeter with internal resistance $90\,\Omega$ reads $1.85\,A$ when connected in a circuit containing a battery and two resistors $700\,\Omega$ and $410\,\Omega$ in series. The actual current in the circuit (without the ammeter) will be:

$A$ $36 \,\Omega$ galvanometer is shunted by a resistance of $4 \,\Omega$. The percentage of the total current that passes through the galvanometer is (in $\%$)

$A$ galvanometer may be converted into an ammeter or a voltmeter. In which of the following cases will the resistance of the device so obtained be the largest?

$A$ galvanometer of resistance $G$ is shunted with a resistance of $10 \%$ of $G$. The part of the total current that flows through the galvanometer is