$A$ galvanometer having a resistance $G$ and current $I_g$ flowing in it,produces full-scale deflection. If $S_1$ is the value of the shunt which converts it into an ammeter of range $0-I$ and $S_2$ is the value of the shunt for range $0-8I$,then the ratio $\frac{S_1}{S_2}$ will be:

$A$ 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:

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter for reading up to $10\,A$,it is necessary to put a resistance of:

In the moving coil galvanometer,if the number of turns in the coil is doubled,then the current sensitivity . . . . . . and the voltage sensitivity . . . . . . .

Assertion: Voltmeter is connected in parallel with the circuit.
Reason: Resistance of a voltmeter is very large.

The magnets of two suspended coil galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side $a$ and that of the second one is circular of radius $\frac{a}{\sqrt{\pi}}.$ When the same current is passed through the coils,the ratio of the torque experienced by the first coil to that experienced by the second one is