The sensitivity of a moving coil galvanometer is $s$. If a shunt of $\frac{1}{8}$ of the resistance of the galvanometer is connected to the moving coil galvanometer,what will its new sensitivity be?

$A$ galvanometer of resistance $G$ can be converted into a voltmeter of range $V$ by connecting a resistance $R$ in series with it. The series resistance required to change its range to $\frac{V}{3}$ is

$A$ moving coil galvanometer has $50$ turns and each turn has an area $2 \times 10^{-4} \ m^2$. The magnetic field produced by the magnet inside the galvanometer is $0.02 \ T$. The torsional constant of the suspension wire is $10^{-4} \ N \ m \ rad^{-1}$. When a current flows through the galvanometer,a full-scale deflection occurs if the coil rotates by $0.2 \ rad$. The resistance of the coil of the galvanometer is $50 \ \Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range $0-1.0 \ A$. For this purpose,a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance,in ohms,is:

$A$ moving coil galvanometer has $48$ turns and the area of the coil is $4 \times 10^{-2} \, m^2$. If the magnetic field is $0.2 \, T$,then to increase the current sensitivity by $25\%$ without changing the area $(A)$ and the magnetic field $(B)$,the number of turns should become:

$A$ $72 \; \Omega$ galvanometer is shunted by a resistance of $8 \; \Omega$. The percentage of the total current which passes through the galvanometer is $.....$

In an ammeter,$5 \%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,the resistance of the ammeter will be: