$A$ 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 ............... $\Omega$.

Potentiometer measures the potential difference more accurately than a voltmeter because

The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \text{ m}$ and its area of cross-section is $8 \times 10^{-6} \text{ m}^2$. If $0.2 \text{ A}$ current is flowing through the wire, the potential gradient of the wire is:

$A$ potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery,used across the potentiometer wire,has an $EMF$ of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R$,connected across the given cell,has values of $(i)$ infinity and $(ii)$ $9.5\,\Omega$,the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m$,respectively. The value of internal resistance of the cell is ............... $\Omega$.

In the given figure,a battery $E$ is balanced on $55 \, cm$ length of a potentiometer wire. When a resistance of $10 \, \Omega$ is connected in parallel with the battery,it balances on $50 \, cm$ length of the potentiometer wire. The internal resistance $r$ of the battery is ............. $\Omega$.

In the given figure,the potentiometer wire $AB$ has a resistance of $5 \, \Omega$ and a length of $10 \, m$. The balancing length $AM$ for the $emf$ of $0.4 \, V$ is ............... $m$.