Two wires A and B with lengths in the ratio of `3:1` diameters in the ratio of `1:2` and resistivities in the ratio of `1:20` are joined in parallel with a source of emf. 2V. Ratio of the
`(R_(1))/(R_(2))` is:

Options

A. `5:2`
B. `2:5`
C. `5:3`
D. `3:5`

Explanation

To find the ratio of the resistances \( R_1 \) and \( R_2 \) of two wires A and B, we can use the formula for resistance:

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

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 1: Define the parameters

Let:
- Length of wire A, \( L_1 = 3x \)
- Length of wire B, \( L_2 = x \)
- Diameter of wire A, \( d_1 = y \)
- Diameter of wire B, \( d_2 = 2y \)
- Resistivity of wire A, \( \rho_1 = z \)
- Resistivity of wire B, \( \rho_2 = 20z \)

### Step 2: Calculate the cross-sectional area

The cross-sectional area \( A \) of a wire is given by:

\[ A = \frac{\pi d^2}{4} \]

So, for wire A:

\[ A_1 = \frac{\pi (d_1)^2}{4} = \frac{\pi (y)^2}{4} \]

And for wire B:

\[ A_2 = \frac{\pi (d_2)^2}{4} = \frac{\pi (2y)^2}{4} = \frac{\pi (4y^2)}{4} = \pi y^2 \]

### Step 3: Write the expressions for resistance

Now, we can write the resistances \( R_1 \) and \( R_2 \):

\[ R_1 = \frac{\rho_1 L_1}{A_1} = \frac{z \cdot 3x}{\frac{\pi (y)^2}{4}} = \frac{12zx}{\pi y^2} \]

\[ R_2 = \frac{\rho_2 L_2}{A_2} = \frac{20z \cdot x}{\pi y^2} = \frac{20zx}{\pi y^2} \]

### Step 4: Calculate the ratio of resistances

Now, we can find the ratio \( \frac{R_1}{R_2} \):

\[ \frac{R_1}{R_2} = \frac{\frac{12zx}{\pi y^2}}{\frac{20zx}{\pi y^2}} = \frac{12}{20} = \frac{3}{5} \]

### Conclusion

Thus, the ratio of the resistances \( \frac{R_1}{R_2} \) is:

\[ \frac{R_1}{R_2} = \frac{3}{5} \]