A wire of length 100 cm is connected to a cell of emf 2 V and negligible internal resistance. The resistance of the wire is `3 Omega`. The additional resistance required to produce a potential drop of 1 milli volt per cm is

Options

A. `60Omega`
B. `47Omega`
C. `57Omega`
D. `35Omega`

Answer

To solve the problem, we need to determine the additional resistance required to produce a potential drop of 1 millivolt per centimeter across a wire of length 100 cm and resistance 3 ohms, when connected to a 2 V battery.

### Step-by-Step Solution:

1. **Identify the total potential drop required across the wire:**
- The wire is 100 cm long, and we want a potential drop of 1 mV per cm.
- Total potential drop across the wire = 100 cm * 1 mV/cm = 100 mV = 0.1 V.

2. **Use Ohm's Law to find the current through the wire:**
- The resistance of the wire is given as 3 ohms.
- According to Ohm's Law, V = I * R.
- Here, we want to find the current (I) that produces a potential drop of 0.1 V across the wire:
\[ I = \frac{V}{R} = \frac{0.1 \, \text{V}}{3 \, \Omega} = \frac{1}{30} \, \text{A}. \]

3. **Set up the equation for the total circuit:**
- The total voltage from the battery is 2 V.
- The total resistance in the circuit is the sum of the wire resistance and the additional resistance (R):
\[ V = I \cdot (R + 3). \]
- Substituting the values we have:
\[ 2 = \left(\frac{1}{30}\right) \cdot (R + 3). \]

4. **Solve for the additional resistance (R):**
- Multiply both sides by 30 to eliminate the fraction:
\[ 60 = R + 3. \]
- Rearranging gives:
\[ R = 60 - 3 = 57 \, \Omega. \]

5. **Conclusion:**
- The additional resistance required to produce a potential drop of 1 mV per cm is **57 ohms**.