Figure shows an arrangement to measure the emf $\mathrm{\epsilon }$and internal resistance $r$ of a battery. The voltmeter has a very high resistance and the ammeter has a very small resistance. The voltmeter reads $1.52 \mathrm{V}$ when the switch $S$ is open. When the switch is closed the voltmeter reading drops to $1.45 \mathrm{V}$ and the ammeter reads $1.0 \mathrm{A}.$ The internal resistance of the battery in $\mathrm{m\Omega }$ will be?

 

In the circuit shown, reading of the voltmeter connected across $400 \mathrm{\Omega }$ resistance is $60 \mathrm{V}.$ If it is connected across $300 \mathrm{\Omega }$ resistance then reading in volt will be

In the given circuit the ammeter $A_1 \text{and} A_2$ are ideal and the ammeter $A_3$ has a resistance of $1.9 \times {10}^{–3} \mathrm{\Omega }$. If sum of readings of all three meters is given by $\left(\frac{2n}{27}\right)$ Ampere the value of $n$ will be. 

The cross section area and length of a cylindrical conductor are $A$ and $l$ respectively and it is connected with a voltage source $V_0$. The conductivity varies as$\mathrm{\sigma }={\mathrm{\sigma }}_0\frac{\mathit{l}}{\mathit{x}}$ where $\mathrm{x} (0 <\mathit{ }x\mathit{ }<\mathit{ }l)$ is the distance along the axis of the cylinder from one of its ends as shown in the figure. Choose the correct option.