The resistance of an ideal ammeter is

Two moving coil meters,$M_{1}$ and $M_{2}$ have the following particulars:
$R_{1}=10 \,\Omega, \quad N_{1}=30$
$A_{1}=3.6 \times 10^{-3} \,m^{2}, \quad B_{1}=0.25 \,T$
$R_{2}=14 \,\Omega, \quad N_{2}=42$
$A_{2}=1.8 \times 10^{-3} \,m^{2}, \quad B_{2}=0.50 \,T$
(The spring constants are identical for the two meters). Determine the ratio of
$(a)$ current sensitivity and
$(b)$ voltage sensitivity of $M_{2}$ and $M_{1}$.

$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is:

Explain: "Increasing the current sensitivity may not necessarily increase the voltage sensitivity".

$A$ galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of up to $10 \ V$. The value of resistance required to convert the galvanometer into an ammeter to read up to $10 \ A$ is $x \times 10^{-2} \Omega$. The value of $x$ is: