$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.

$A$ potentiometer is an ideal device for measuring potential difference because

In a potentiometer,a wire of length $10 \ m$ having resistance $50 \ \Omega$ is used. $A$ battery of $5 \ V$ and a resistor of $450 \ \Omega$ are connected in series to the wire. If an unknown battery of emf $E$ balances the potentiometer at $450 \ cm$,then the value of $E$ is (in $V$)

$A$ cell can be balanced against $110 \ cm$ and $100 \ cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \ \Omega$. Its internal resistance is

The sliding contact of a potentiometer is in the middle of the potentiometer wire having a total resistance $R_p = 1 \Omega$,as shown in the figure. An external resistance of $R_e = 2 \Omega$ is connected via the sliding contact. Find the total current drawn from the $0.9 \text{ V}$ battery. (in $\text{ A}$)

$A$ resistance of $4\,\Omega$ and a wire of length $5\,m$ and resistance $5\,\Omega$ are joined in series and connected to a cell of $e.m.f.$ $10\,V$ and internal resistance $1\,\Omega$. $A$ parallel combination of two identical cells is balanced across $300\,cm$ of the wire. The $e.m.f.$ $E$ of each cell is ........... $V$.