$A$ galvanometer of resistance $10 \Omega$ changes its range from $1 \text{ mA}$ to $101 \text{ mA}$ when a resistive wire is connected in parallel. If the resistivity of the material of the wire and its area of cross-section are $1 \times 10^{-6} \Omega \text{ m}$ and $1 \text{ mm}^2$ respectively,then the length of the wire is (in $cm$)

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 ............... $\Omega$.

The maximum current that can be measured by a galvanometer of resistance $40 \,\Omega$ is $10 \,mA$. It is converted into a voltmeter that can read up to $50 \,V$. The resistance to be connected in series with the galvanometer is ... (in $\Omega$)

If the number of turns in the coil of a galvanometer is decreased,then the resistance of the galvanometer:

$A$ galvanometer,having a resistance of $50 \Omega$,gives a full scale deflection for a current of $0.05 \text{ A}$. The length in metre of a resistance wire of area of cross-section $2.97 \times 10^{-2} \text{ cm}^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \text{ A}$ current is: (Specific resistance of the wire $= 5 \times 10^{-7} \Omega\text{-m}$)

The galvanometer has a resistance of $1.8 \Omega$. Calculate the value of shunt to increase the range of galvanometer by $10$ times. (in $\Omega$)