Shunt wire should be . . . . . . .

To reduce the current flowing through a coil of resistance $90\, \Omega$ by $90\%$,a resistor of what value in $\Omega$ must be connected in parallel?

$A$ 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 $Omega$)

To find the resistance of a galvanometer by the half-deflection method,the experimental data obtained are given in the table below:

$S. No.$	Resistance $R \ (\Omega)$	Deflection $(\theta)$	Shunt $S \ (\Omega)$	Half deflection $(\theta / 2)$	Galvanometer resistance $(G)$
$1$	$3300$	$30$	$80$	$15$	$G_1$
$2$	$5000$	$20$	$80$	$10$	$G_2$

From the above data,the galvanometer resistance $G$ will be near to: (in $\Omega$)

In the experiment to determine the galvanometer resistance by the half-deflection method,the plot of $\frac{1}{\theta}$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is .............. $\times 10^{-1} \text{ A/division}$. [The source has emf $E = 2 \text{ V}$]

$A$ galvanometer of resistance $240 \Omega$ allows only $4 \%$ of the main current to pass through it after connecting a shunt resistance. The value of the shunt resistance is (in $Omega$)