The emf of a cell having internal resistance $1 \Omega$ is balanced against a length of $330 \mathrm{~cm}$ on a potentiometer wire. When an external resistance of $2 \Omega$ is connected across the cell, the balancing length will be

• A

  $115 \mathrm{~cm}$
• B

  $332 \mathrm{~cm}$
• Correct answer
  C

  $220 \mathrm{~cm}$
• D

  $330 \mathrm{~cm}$

Solution

For case - I $\begin{aligned} & \Rightarrow \varepsilon_s=\phi \times 330 \\ & \Rightarrow \phi=\frac{\varepsilon_s}{330}\left(\frac{V}{\mathrm{~cm}}\right) \ldots\end{aligned}$ For case - II $\begin{aligned} & \Rightarrow i \times 2=\phi I_2 \\ & \Rightarrow \frac{\varepsilon_s}{3} \times 2=\frac{\varepsilon_s}{330} \times I_2 \text { [From equation 1] } \\ & \Rightarrow I_2=220 \mathrm{~cm}\end{aligned}$

Hence, the correct answer is option: 'C'.