A wire of length "10cm" and radius `sqrt(7)times10^(-4)m` is connected across the right gap of a meter bridge.When a resistance of `4.5Omega` is connected on the left gap by using a resistance box,the balance length is found to be at "60cm" from the left end.If the resistivity of the wire is `R times10^(-7), Omega m` ,then value of "R" is :

Options

A. 63
B. 70
C. 35
D. 66

Answer

To solve the problem step by step, we will follow the principles of a meter bridge and the formula for resistance in terms of resistivity, length, and cross-sectional area.

### Step 1: Understand the Meter Bridge Setup
In a meter bridge, when the bridge is balanced, the ratio of the resistances is equal to the ratio of the lengths from the point of balance to the ends of the bridge.

Given:
- Resistance on the left (R1) = 4.5 Ω
- Balance length from the left end (L1) = 60 cm
- Length of the meter bridge (total length) = 100 cm
- Therefore, length on the right (L2) = 100 cm - 60 cm = 40 cm

### Step 2: Apply the Balance Condition
Using the balance condition of the meter bridge:
\[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \]
Substituting the known values:
\[ \frac{4.5}{R} = \frac{60}{40} \]
Simplifying the right side:
\[ \frac{4.5}{R} = \frac{3}{2} \]

### Step 3: Solve for R
Cross-multiplying gives:
\[ 4.5 \times 2 = 3R \]
\[ 9 = 3R \]
\[ R = \frac{9}{3} = 3 \, \Omega \]

### Step 4: Calculate the Resistivity
The resistance of a wire can be expressed as:
\[ R = \rho \frac{L}{A} \]
Where:
- \( R \) = resistance (3 Ω)
- \( \rho \) = resistivity (in the form \( R \times 10^{-7} \, \Omega \cdot m \))
- \( L \) = length of the wire (10 cm = 0.1 m)
- \( A \) = cross-sectional area of the wire

The cross-sectional area \( A \) of the wire can be calculated using the radius:
\[ A = \pi r^2 \]
Given radius \( r = \sqrt{7} \times 10^{-4} \, m \):
\[ A = \pi \left(\sqrt{7} \times 10^{-4}\right)^2 = \pi \times 7 \times 10^{-8} = 7\pi \times 10^{-8} \, m^2 \]

### Step 5: Substitute Values into the Resistance Formula
Now substituting the values into the resistance formula:
\[ 3 = \rho \frac{0.1}{7\pi \times 10^{-8}} \]
Rearranging for \( \rho \):
\[ \rho = 3 \cdot \frac{7\pi \times 10^{-8}}{0.1} \]
\[ \rho = 21\pi \times 10^{-7} \, \Omega \cdot m \]

### Step 6: Determine the Value of R
Since \( \rho = R \times 10^{-7} \, \Omega \cdot m \), we can equate:
\[ R \times 10^{-7} = 21\pi \times 10^{-7} \]
Thus, \( R = 21\pi \).

### Step 7: Calculate the Numerical Value
Using \( \pi \approx 3.14 \):
\[ R \approx 21 \times 3.14 \approx 66 \]

### Conclusion
The value of \( R \) is approximately 66.