To convert a galvanometer into a voltmeter,one should connect a

If the number of turns in the coil of a moving coil galvanometer decreases, the resistance of the galvanometer,

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

$A$ voltmeter of resistance $R$ is to be used to measure the potential difference across a resistance $r$. In which case will the reading be nearest to the actual potential difference in the absence of the voltmeter?

$A$ student uses the resistance of a known resistor $(1 \,\Omega)$ to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be $1 \times 10^3 \,\Omega$ in circuit $(a)$ and $0.999 \,\Omega$ in circuit $(b)$. From these measurements,the resistance (in $\Omega$) of the voltmeter and ammeter are found to be close to

If $n$ represents the actual number of deflections in a converted galvanometer of resistance $G$ and shunt resistance $S$,then the total current $I$ when its figure of merit is $K$ will be: