In the circuit shown below,the reading of the voltmeter $V$ is ........... (in $, V$)

$A$ cell of $emf$ $E$ and internal resistance $r$ is connected in series with an external resistance $nr$. Then,the ratio of the terminal potential difference to $emf$ is

$A$ voltmeter of resistance $998\ \Omega$ is connected to a cell of $emf$ $2\ V$ and internal resistance $2\ \Omega$. The error in the measurement of $emf$ will be:

Assertion : In a simple battery circuit,the point of the lowest potential is the negative terminal of the battery.
Reason : The current flows towards the point of the higher potential,as it does in such a circuit from the negative to the positive terminal.

$A$ battery of internal resistance $4 \, \Omega$ is connected to the network of resistors as shown. In order for the maximum power to be delivered to the network, the value of $R$ in $\Omega$ should be:

$A$ resistance network is connected to a battery as shown in the figure below. If the internal resistance of the battery is $5 \ \Omega$,then the value of $R$ (in $\Omega$) for maximum power delivered to the network is