If an ammeter is to be used in place of a galvanometer,then we must connect:

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

$A$ galvanometer coil has a resistance $90\, \Omega$ and full-scale deflection current $10\, mA$. $A$ $910\, \Omega$ resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.1\, V$,the number of divisions on its scale is:

For designing a voltmeter of range $50\,V$ and an ammeter of range $10\,mA$ using a galvanometer which has a coil of resistance $54\,\Omega$ showing a full scale deflection for $1\,mA$ as in figure.
$(A)$ for voltmeter $R \approx 50\,k\Omega$
$(B)$ for ammeter $r \approx 0.2\,\Omega$
$(C)$ for ammeter $r \approx 6\,\Omega$
$(D)$ for voltmeter $R \approx 5\,k \Omega$
$(E)$ for voltmeter $R \approx 500 \Omega$
Choose the correct answer from the options given below

$A$ galvanometer of resistance $100 \Omega$ gives full scale deflection for a current of $1 \text{ mA}$. It is converted into an ammeter of range $0 - 10 \text{ A}$. The shunt required is: (in $\text{ }\Omega$)

$A$ galvanometer gives full scale deflection with $0.006 \ A$ current. By connecting it to a $4990 \ \Omega$ resistance,it can be converted into a voltmeter of range $0-30 \ V$. If connected to a $\frac{2n}{249} \ \Omega$ resistance,it becomes an ammeter of range $0-1.5 \ A$. The value of $n$ is: