Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625 \, cm$. If the terminals of $E_1$ are reversed,the balancing length obtained is $125 \, cm$. Given $E_2 > E_1$,the ratio $E_1 : E_2$ will be

With a potentiometer,null points were obtained at $140 \, cm$ and $180 \, cm$ with cells of $emf$ $1.1 \, V$ and one unknown $X \, V$. The unknown $emf$ is .............. $V$.

$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

An ideal battery of $4\, V$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1\, m$ and resistance $5\,\Omega$. The value of $R$,to give a potential difference of $5\, mV$ across $10\, cm$ of potentiometer wire,is: ................ $\Omega$

$A$ potentiometer wire has length $4\, m$ and resistance $8\, \Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\, V$,so as to get a potential gradient of $1\, mV$ per $cm$ on the wire is ............. $\Omega$.

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 $\text{ A}$)