A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V as shown in the figure. When a cell of 1.5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this cell and using a cell of unknown emf, the balance point shifts to 80 cm.
(i) Calculate unknown emf of the cell.
(ii) Explain with reason, whether the circuit works, if the driver cell is replaced with a cell of emf 1V.
(iii) Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer.

Given,
Length of the potentiometer wire, l = 1 m
Emf of the driver cell , E = 3 V
Emf of the cell in the secondary circuit, E = 1.5 V
Balance point obtained, l₁ = 60 cm= 0.60 m
Now,
On replacing the cell by a cell of unknown emf, the balance point obtained is at, l₂ = 80 cm = 0.80 m
(i) By using the formula, 

             $$\begin{gathered} \boxed{\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2} = \frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}} \\ \Rightarrow {\mathrm{E}}_2 = \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1} \times {\mathrm{E}}_1 \end{gathered}$$ 

$\therefore$ ${\mathrm{E}}_2 = \frac{0.80}{0.60}\times 1.5 = \frac{80 \times 1.5}{60} = 2 \mathrm{V}$ 
Hence, the unknown emf is found to be 2V.

(ii) The circuit will not work.

Reason: Because there will be smaller fall of potential across the potentiometer wire as compared to the emf of the cell in secondary circuit to be determined. Hence, the balance point will not be obtained on the potentiometer wire. 
Thus, the condition for obtaining the balance point is that, the emf of the driver cell should be greater than the emf of the cell to be determined.

(iii) High resistance R, used in the secondary circuit does not affect the balance point because, at balance point the galvanometer shows no deflection implying the absence of current in this condition.