$A$ $10\,V$ battery with internal resistance $1\,\Omega$ and a $15\,V$ battery with internal resistance $0.6\,\Omega$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to ................ $V$.

Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:

Why is the combination of cells done? Write its methods.

What is called the internal resistance of a cell?

Two cells of emfs $1$ $V$ and $2$ $V$ and internal resistances $2 \Omega$ and $1 \Omega$ respectively are connected in parallel and provide a current of $1$ $A$ through an external resistance. If the polarity of one cell is reversed, the current through the external resistance becomes $\frac{\alpha}{5}$ $A$. The value of $\alpha$ is . . . . . . .

$100$ cells, each of $e.m.f.$ $5\, V$ and internal resistance $1\, \Omega$, are to be arranged to produce maximum current in a $25\, \Omega$ external resistance. Each row must contain an equal number of cells. The number of rows should be: