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

$A$ galvanometer of resistance $20 \Omega$ gives a deflection of $5$ divisions when $1 \text{ mA}$ current flows through it. The galvanometer scale has $50$ divisions. To convert the galvanometer into a voltmeter of range $25 \text{ V}$,we should connect a resistance of

Match the terms given in List-$I$ with suitable options provided in List-$II$ regarding the experiment to calculate the resistance and figure of merit of a galvanometer using the half-deflection method:

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(SI)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using low resistance	$R$. in shunt/parallel
$D$. High resistance box	$S$. current per unit deflection
	$T$. deflection per unit current

The percentage error in the reading of the voltmeter in the figure shown here is nearly ............ $\%$

$A$ shunt of resistance $1\ \Omega$ is connected across a galvanometer of $120\ \Omega$ resistance. $A$ current of $5.5\ A$ flows through the circuit,and the galvanometer shows full-scale deflection. What is the current that would cause full-scale deflection in the galvanometer in the absence of the shunt (in $A$)?

$A$ $50\, \Omega$ galvanometer gets full-scale deflection when a current of $0.01\, A$ passes through the coil. When it is converted to a $10\, A$ ammeter,the shunt resistance is ........... $\Omega$.