Consider the circuit shown in the figure. The value of current 'I' is

A milliammeter of resistance $40 \Omega$ has a range $0-30 \mathrm{~mA}$. What will be the resistance used in series to convert it into voltmeter of range $0-15 \mathrm{~V}$ ?

Two cells having unknown emfs $E_1$ and $E_2$ $\left(E_1>E_2\right)$ are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell $E_2$ is connected, so as to oppose cell $E_1$, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells $\left(\frac{E_1}{E_2}\right)$ is

A 10 m long wire of resistance $20 \Omega$ is connected in series with a battery of emf 3 V and a resistance of $10 \Omega$. The potential gradient along the wire in $\mathrm{V} / \mathrm{m}$ is