If the current is doubled,the deflection is also doubled in

How can we convert a galvanometer into a voltmeter?

$A$ 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 $= 5 \times 10^{-7} \Omega\text{-m}$)

For a moving coil galvanometer,the deflection in the coil is $0.05\,rad$ when a current of $10\,mA$ is passed through it. If the torsional constant of the suspension wire is $4.0 \times 10^{-5}\,Nm\,rad^{-1}$,the magnetic field is $0.01\,T$,and the number of turns in the coil is $200$,the area of each turn (in $cm^2$) is:

The sensitivity of a galvanometer of resistance $171 \, \Omega$ becomes $1/20$ times when we use a shunt of $x \, \Omega$. Find $x$ ................. $\Omega$.

Derive the expression for the shunt resistance required to increase the current-measuring capacity of a galvanometer by a factor of $n$.