Current sensitivities of two galvanometers $G_1$ and $G_2$ of resistances $100 \ \Omega$ and $50 \ \Omega$ are $10^8 \ \text{div/A}$ and $0.5 \times 10^5 \ \text{div/A}$ respectively. The galvanometer in which the voltage sensitivity is more is

The coil in a moving coil galvanometer experiences a torque proportional to the current passed through it. If a steady current $i$ is passed through it,the deflection of the coil is found to be $90^o$. Now,the steady current is switched off and a charge $Q$ is suddenly passed through the coil. If the coil has $N$ turns of area $A$ each and its moment of inertia is $I$ about the axis it is going to rotate,then the maximum angle through which it deflects upon passing $Q$ is:

$A$ galvanometer has a resistance of $25 \, \Omega$ and a maximum of $0.01 \, A$ current can be passed through it. In order to change it into an ammeter of range $10 \, A$, the shunt resistance required is

$A$ moving coil galvanometer has $100$ turns and each turn has an area of $2.0 \,cm^2$. The magnetic field produced by the magnet is $0.01 \,T$ and the deflection in the coil is $0.05$ radian when a current of $10 \,mA$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5} \,N-m / rad$. The value of $x$ is . . . . . . .

To find the resistance of a galvanometer by the half-deflection method,the following circuit is used with resistances $R_1 = 9970\,\Omega$,$R_2 = 30\,\Omega$,and $R_3 = 0\,\Omega$. The deflection in the galvanometer is $d$. With $R_3 = 107\,\Omega$,the deflection changes to $\frac{d}{2}$. The galvanometer resistance is approximately ............... $\Omega$.

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