A cell of e.m.f 1.5V and negligible internal resistance is connected in series with a potential meter of length 10 m and total resistance `20 Omega`. What resistance should be introduced in the resistance box such that the potential drop across the potentiometer is one microvolt per cm of the wire?

Answer

To solve the problem, we need to determine the resistance that should be introduced in the resistance box so that the potential drop across the potentiometer is one microvolt (1 µV) per centimeter of the wire.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- EMF of the cell (E) = 1.5 V
- Length of the potentiometer wire (L) = 10 m = 1000 cm
- Total resistance (R) = 20 Ω
- Desired potential gradient (V_g) = 1 µV/cm = 1 × 10^(-6) V/cm

2. **Calculate the Total Potential Drop Across the Potentiometer:**
- The total potential drop (V_total) across the potentiometer wire can be calculated as:
\[ V_{total} = V_g \times L = (1 \times 10^{-6} \, \text{V/cm}) \times (1000 \, \text{cm}) = 1 \times 10^{-3} \, \text{V} = 0.001 \, \text{V} \]

3. **Using Ohm's Law to Find Current (I):**
- The potential drop across the total resistance (R + R_e) is given by Ohm's Law:
\[ V_{total} = I \times R_{eq} \]
- Where \( R_{eq} = R + R_e \) (R is the known resistance and \( R_e \) is the resistance we need to find).
- Rearranging gives:
\[ I = \frac{V_{total}}{R_{eq}} = \frac{0.001}{R + R_e} \]

4. **Current through the Circuit:**
- We can also express the current in terms of the EMF and total resistance:
\[ I = \frac{E}{R_{eq}} = \frac{1.5}{R + R_e} \]

5. **Setting the Two Expressions for Current Equal:**
- Equating the two expressions for current:
\[ \frac{0.001}{R + R_e} = \frac{1.5}{R + R_e} \]
- This simplifies to:
\[ 0.001 = 1.5 \times \frac{R}{R + R_e} \]

6. **Substituting the Known Resistance (R):**
- Substitute \( R = 20 \, \Omega \):
\[ 0.001 = 1.5 \times \frac{20}{20 + R_e} \]

7. **Solving for \( R_e \):**
- Rearranging gives:
\[ 0.001(20 + R_e) = 30 \implies 0.02 + 0.001R_e = 30 \]
- Thus:
\[ 0.001R_e = 30 - 0.02 = 29.98 \implies R_e = \frac{29.98}{0.001} = 29980 \, \Omega \]

### Final Answer:
The resistance that should be introduced in the resistance box is approximately **29980 Ω**.