$A$ cell balances against a length of $150 \ cm$ on a potentiometer wire when it is shunted by a resistance of $5 \ \Omega$. But when it is shunted by a resistance of $10 \ \Omega$,then the balancing length increases by $25 \ cm$. The balancing length when the cell is in an open circuit is: (in $cm$)

As shown in the figure,a potentiometer wire of resistance $20\,\Omega$ and length $300\,cm$ is connected with a resistance box ($R$.$B$.) and a standard cell of emf $4\,V$. For a resistance '$R$' of the resistance box introduced into the circuit,the null point for a cell of $20\,mV$ is found to be $60\,cm$. The value of '$R$' is $.....\Omega$

$A$ cell,shunted by an $8 \; \Omega$ resistance,is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by a $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega$.

When a cell of $E$.$M$.$F$. '$E_1$' is connected to a potentiometer wire, the balancing length is '$l_1$'. Another cell of $E$.$M$.$F$. '$E_2$' $(E_1 > E_2)$ is connected along with $E_1$ such that the two cells oppose each other, and the balancing length is '$l_2$'. The ratio $E_1 : E_2$ is

The figure shows a simple potentiometer circuit for measuring a small $e.m.f.$ produced by a thermocouple. The potentiometer wire $PQ$ has a resistance of $5 \,\Omega$ and the driver cell has an $e.m.f.$ of $2 \, V$. If a balance point is obtained at $0.600 \, m$ along $PQ$ when measuring an $e.m.f.$ of $6.00 \, mV$,what is the value of resistance $R$ in $\Omega$?

In a potentiometer arrangement,a cell of $emf$ $1.25\; V$ gives a balance point at $35.0\; cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\; cm ,$ what is the $emf$ of the second cell in $V$?