$A$ potentiometer wire has length $L$. For a given cell of emf $E$,the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of the potentiometer wire is increased by $50 \%$,then for the same cell,the balance point is obtained at length

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

In the determination of the internal resistance of a cell with a potentiometer,the error in the measurement of the balancing length is $\pm 1 \text{ mm}$. When the cell alone is connected in the circuit,the balancing length is obtained at $60 \text{ cm}$ and when the cell is shunted with a resistance of $10 \Omega \pm 2 \%$,the balancing length is obtained at $50 \text{ cm}$. The error in the determination of the internal resistance of the cell is (in $\%$)

If the length of the potentiometer wire is increased by keeping the potential difference across the wire constant,then:

In an experiment with a potentiometer,$V_B = 10 \, V$ and the variable resistance is adjusted to $R = 50 \, \Omega$ (figure). $A$ student wanting to measure the voltage $E_1$ of a battery (approx. $8 \, V$) finds no null point. He then diminishes $R$ to $10 \, \Omega$ and is able to locate the null point on the last $(4^{th})$ segment of the potentiometer wire. Find the resistance of the potentiometer wire and the potential drop per unit length.

Resistance of $100 \, cm$ long potentiometer wire is $10 \, \Omega$. It is connected to a battery of $2 \, V$ and a resistance $R$ in series. $A$ source of $10 \, mV$ gives a null point at $40 \, cm$ length. The value of external resistance $R$ is ........... $\Omega$.