Two uniform wires `A` and `B` are of the same total metal and have equal masses. The radius of wire `A` is twice that of wire `B`. The total resistance of `A` and `B` when connected in parallel is

Options

A. `4Omega` when the resistance of wire A is `4.25Omega`
B. `5Omega` when the resistnace of wire A is `4Omega`
C. `4Omega` when the resistance of wire B is `4.25Omega`
D. `5Omega` when the resistance of wire B is `4Omega`

Answer

To solve the problem, we need to find the total resistance of two uniform wires A and B connected in parallel. Given that both wires have the same mass and are made of the same material, we can derive their resistances based on their dimensions.

### Step 1: Understand the relationship between radius and resistance
The resistance \( R \) of a wire is given by the formula:

\[ R = \frac{\rho L}{A} \]

where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

The cross-sectional area \( A \) of a wire with radius \( r \) is:

\[ A = \pi r^2 \]

### Step 2: Define the variables for wires A and B
Let:
- The radius of wire B be \( r_B \).
- The radius of wire A be \( r_A = 2r_B \) (since the radius of wire A is twice that of wire B).

### Step 3: Express the volumes of the wires
Since both wires have the same mass and are made of the same material, we can express their volumes in terms of their dimensions:

\[ \text{Volume} = A \cdot L \]

For wire A:

\[ V_A = \pi (r_A^2) L_A = \pi (2r_B)^2 L_A = 4\pi r_B^2 L_A \]

For wire B:

\[ V_B = \pi (r_B^2) L_B \]

Since the masses are equal and the density is the same, we have:

\[ \rho V_A = \rho V_B \implies V_A = V_B \]

Thus:

\[ 4\pi r_B^2 L_A = \pi r_B^2 L_B \]

Cancelling \( \pi r_B^2 \) (assuming \( r_B \neq 0 \)) gives:

\[ 4L_A = L_B \implies L_B = 4L_A \]

### Step 4: Calculate the resistances of wires A and B
Now we can express the resistances of both wires using their lengths:

For wire A:

\[ R_A = \frac{\rho L_A}{\pi (2r_B)^2} = \frac{\rho L_A}{4\pi r_B^2} \]

For wire B:

\[ R_B = \frac{\rho L_B}{\pi r_B^2} = \frac{\rho (4L_A)}{\pi r_B^2} = \frac{4\rho L_A}{\pi r_B^2} \]

### Step 5: Relate the resistances
Now we can find the ratio of the resistances:

\[ \frac{R_B}{R_A} = \frac{\frac{4\rho L_A}{\pi r_B^2}}{\frac{\rho L_A}{4\pi r_B^2}} = \frac{4}{\frac{1}{4}} = 16 \]

Thus, we can say:

\[ R_B = 16 R_A \]

### Step 6: Calculate the equivalent resistance in parallel
The equivalent resistance \( R_p \) of two resistances \( R_A \) and \( R_B \) in parallel is given by:

\[ \frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{R_B} \]

Substituting \( R_B = 16 R_A \):

\[ \frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{16 R_A} = \frac{16 + 1}{16 R_A} = \frac{17}{16 R_A} \]

Thus:

\[ R_p = \frac{16 R_A}{17} \]

### Step 7: Conclusion
The total resistance of wires A and B when connected in parallel is:

\[ R_p = \frac{16}{17} R_A \]