To solve this problem, we begin by understanding the scenario: A wire of length \(100~\mathrm{cm}\) is connected to a cell with an emf of \(2~\mathrm{V}\) and negligible internal resistance. The wire has a resistance of \(3~\Omega\). We need to find the additional resistance that is required to achieve a potential difference of \(1~\mathrm{mV/cm}\) across the wire.
- First, calculate the total potential difference needed across the 100 cm wire:
- Potential difference per unit length: \(1~\mathrm{mV/cm}\).
- Total length of wire = \(100~\mathrm{cm}\).
- Total potential difference required, \(V = 1~\mathrm{mV/cm} \times 100~\mathrm{cm} = 100~\mathrm{mV} = 0.1~\mathrm{V}\).
- Determine the total resistance needed to achieve this potential difference:
- Ohm's Law states: \(V = I \cdot R\).
- The total emf available is \(2~\mathrm{V}\), and we need \(0.1~\mathrm{V}\) across the wire, leaving \(2~\mathrm{V} - 0.1~\mathrm{V} = 1.9~\mathrm{V}\) across the additional resistance.
- Find the total current flowing through the circuit:
- The wire carries a current \(I\), given by \(I = \frac{V_{\text{wire}}}{R_{\text{wire}}} = \frac{0.1~\mathrm{V}}{3~\Omega} = \frac{1}{30}~\mathrm{A}\).
- Calculate the additional resistance:
- The potential difference across the total resistance \(R_{\text{total}}\) is \(2~\mathrm{V}\) and the resistance should produce a current \(I = \frac{1}{30}~\mathrm{A}\).
- Using Ohm’s Law for the total circuit: \(R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{2~\mathrm{V}}{\frac{1}{30}~\mathrm{A}} = 60~\Omega\).
- The wire itself contributes \(3~\Omega\), so the additional resistance required is \(R_{\text{add}} = 60~\Omega - 3~\Omega = 57~\Omega\).
Thus, the additional resistance required to achieve the desired potential difference is \(57~\Omega\).