In a metere bridge, the balance length from left end (standard resistance of `1 Omega ` is in theright gap) is found to be 20cm the length of resistance in left gap is 1/2 m and radius is 2mm its specific resistance is

Options

A. `pi xx 10^(-6) ohm -m`
B. `2 pi xx 10^(-6) ohm -m`
C. `(pi)/(2) xx 10^(-6) ohm -m`
D. `3pi xx 10^(-6) ohm -m`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Balancing length from the left end (l1) = 20 cm = 0.2 m
- Length of resistance in the left gap (l) = 0.5 m
- Radius of the wire (r) = 2 mm = 2 × 10^-3 m
- Standard resistance in the right gap (r2) = 1 Ω

### Step 2: Calculate the balancing length from the right end
The total length of the meter bridge is 100 cm (1 m). Thus, the balancing length from the right end (l2) can be calculated as:
\[ l2 = 100 \text{ cm} - l1 = 100 \text{ cm} - 20 \text{ cm} = 80 \text{ cm} = 0.8 \text{ m} \]

### Step 3: Use the meter bridge formula
According to the meter bridge principle:
\[ \frac{r1}{r2} = \frac{l1}{l2} \]
Where:
- \( r1 \) = resistance in the left gap
- \( r2 \) = standard resistance in the right gap

Rearranging the formula to find \( r1 \):
\[ r1 = \frac{l1}{l2} \times r2 \]

### Step 4: Substitute the values
Substituting the known values:
\[ r1 = \frac{0.2 \text{ m}}{0.8 \text{ m}} \times 1 \text{ Ω} = \frac{1}{4} \text{ Ω} = 0.25 \text{ Ω} \]

### Step 5: Calculate the cross-sectional area (A) of the wire
The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (2 \times 10^{-3} \text{ m})^2 = \pi (4 \times 10^{-6} \text{ m}^2) = 4\pi \times 10^{-6} \text{ m}^2 \]

### Step 6: Use the formula for specific resistance (ρ)
The resistance \( r1 \) is also given by the formula:
\[ r1 = \frac{\rho l}{A} \]
Rearranging to find the specific resistance \( \rho \):
\[ \rho = \frac{r1 \cdot A}{l} \]

### Step 7: Substitute the values into the specific resistance formula
Substituting the known values:
\[ \rho = \frac{(0.25 \text{ Ω}) \cdot (4\pi \times 10^{-6} \text{ m}^2)}{0.5 \text{ m}} \]
\[ \rho = \frac{1 \pi \times 10^{-6} \text{ Ω m}}{0.5} = 2\pi \times 10^{-6} \text{ Ω m} \]

### Final Answer
Thus, the specific resistance \( \rho \) is:
\[ \rho = 2\pi \times 10^{-6} \text{ Ω m} \]

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