Two wires have lengths, diameters and specific resistances all in the ratio of `1: 2`. The resistance of the first wire is 10 ohm. Resistance of the second wire in ohm will be

Options

A. `5`
B. `10`
C. `20`
D. infinite

Explanation

To find the resistance of the second wire given the information about the first wire and the ratios of their dimensions, we can follow these steps:

### Step 1: Understand the relationship between resistance, resistivity, length, and area
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Define the ratios for the two wires
Given that the lengths, diameters, and specific resistances of the two wires are in the ratio \( 1:2 \), we can express them as:
- Length of wire 1, \( L_1 = L \)
- Length of wire 2, \( L_2 = 2L \)
- Diameter of wire 1, \( d_1 = d \)
- Diameter of wire 2, \( d_2 = 2d \)
- Resistivity of wire 1, \( \rho_1 = \rho \)
- Resistivity of wire 2, \( \rho_2 = 2\rho \)

### Step 3: Calculate the area of cross-section for both wires
The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]
Thus, we can calculate the areas:
- Area of wire 1, \( A_1 = \frac{\pi d^2}{4} \)
- Area of wire 2, \( A_2 = \frac{\pi (2d)^2}{4} = \frac{\pi (4d^2)}{4} = \pi d^2 \)

### Step 4: Write the resistance formulas for both wires
Using the resistance formula:
- Resistance of wire 1:
\[ R_1 = \frac{\rho_1 L_1}{A_1} = \frac{\rho L}{\frac{\pi d^2}{4}} = \frac{4\rho L}{\pi d^2} \]
- Resistance of wire 2:
\[ R_2 = \frac{\rho_2 L_2}{A_2} = \frac{(2\rho)(2L)}{\pi d^2} = \frac{4\rho L}{\pi d^2} \]

### Step 5: Find the ratio of the resistances
Now we can find the ratio of the resistances:
\[ \frac{R_1}{R_2} = \frac{\frac{4\rho L}{\pi d^2}}{\frac{4\rho L}{\pi d^2}} = 1 \]
Since \( R_1 = 10 \, \Omega \), we can express \( R_2 \):
\[ R_2 = 10 \, \Omega \times \left(\frac{A_1}{A_2}\right) \]
Since \( A_2 = 4A_1 \):
\[ R_2 = 10 \, \Omega \times \frac{1}{4} = 40 \, \Omega \]

### Conclusion
Thus, the resistance of the second wire \( R_2 \) is:
\[ R_2 = 40 \, \Omega \]