Two wires `A` and `B` of same material and same mass have radius `2r` and `r`. If resistance of wire `A` is `34 Omega`, then resistance of `B` will be

Options

A. `544 Omega`
B. `372 Omega`
C. `68 Omega`
D. `17 Omega`

Explanation

To find the resistance of wire B given that the resistance of wire A is 34 ohms, we can follow these steps:

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

### Step 2: Identify the properties of the wires
Both wires A and B are made of the same material and have the same mass. The radius of wire A is \( 2r \) and the radius of wire B is \( r \).

### Step 3: Calculate the cross-sectional areas
The cross-sectional area \( A \) of a wire can be calculated using the formula:
\[ A = \pi r^2 \]
For wire A:
\[ A_A = \pi (2r)^2 = \pi \cdot 4r^2 = 4\pi r^2 \]
For wire B:
\[ A_B = \pi r^2 \]

### Step 4: Relate the volumes of the wires
Since both wires have the same mass and are made of the same material, their volumes must also be equal. The volume \( V \) of a wire is given by:
\[ V = A \cdot L \]
Thus, we can write:
\[ A_A \cdot L_A = A_B \cdot L_B \]
Substituting the areas:
\[ (4\pi r^2) \cdot L_A = (\pi r^2) \cdot L_B \]
Dividing both sides by \( \pi r^2 \):
\[ 4L_A = L_B \quad \Rightarrow \quad \frac{L_A}{L_B} = \frac{1}{4} \]

### Step 5: Substitute into the resistance formula
Now we can express the resistance of wire B in terms of the resistance of wire A:
\[ \frac{R_A}{R_B} = \frac{\rho \frac{L_A}{A_A}}{\rho \frac{L_B}{A_B}} = \frac{L_A}{A_A} \cdot \frac{A_B}{L_B} \]
Substituting the values we have:
\[ \frac{R_A}{R_B} = \frac{L_A}{4\pi r^2} \cdot \frac{\pi r^2}{L_B} \]
Using \( \frac{L_A}{L_B} = \frac{1}{4} \):
\[ \frac{R_A}{R_B} = \frac{1/4}{4} = \frac{1}{16} \]

### Step 6: Calculate the resistance of wire B
Given that \( R_A = 34 \, \Omega \):
\[ \frac{34}{R_B} = \frac{1}{16} \quad \Rightarrow \quad R_B = 34 \times 16 = 544 \, \Omega \]

### Final Answer
The resistance of wire B is:
\[ R_B = 544 \, \Omega \]