Two wires of the same material have resistances in the ratio ` 16 : 81` and lengths in the ratio ` 1 : 4`. Compare their radii of cross-section.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between resistance, length, and cross-sectional 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 of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 2: Set up the ratios for the two wires
Let the two wires be Wire 1 and Wire 2. We know:
- The ratio of their resistances \( R_1 : R_2 = 16 : 81 \)
- The ratio of their lengths \( L_1 : L_2 = 1 : 4 \)

### Step 3: Write the equation for the resistance ratio
Using the resistance formula, we can write:
\[ \frac{R_1}{R_2} = \frac{\rho L_1}{A_1} \cdot \frac{A_2}{\rho L_2} \]
This simplifies to:
\[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \cdot \frac{A_2}{A_1} \]

### Step 4: Substitute the known ratios
Substituting the known ratios into the equation, we have:
\[ \frac{16}{81} = \frac{1}{4} \cdot \frac{A_2}{A_1} \]

### Step 5: Solve for the area ratio
Rearranging the equation gives:
\[ \frac{A_2}{A_1} = \frac{16}{81} \cdot 4 = \frac{64}{81} \]

### Step 6: Relate the area to the radius
The area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
Thus, we can write:
\[ \frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} = \frac{r_2^2}{r_1^2} \]
This leads to:
\[ \frac{r_2^2}{r_1^2} = \frac{64}{81} \]

### Step 7: Take the square root to find the radius ratio
Taking the square root of both sides gives:
\[ \frac{r_2}{r_1} = \frac{8}{9} \]

### Step 8: Find the final ratio of the radii
To find the ratio of \( r_1 \) to \( r_2 \):
\[ \frac{r_1}{r_2} = \frac{9}{8} \]

### Final Result
The ratio of the radii of the two wires is:
\[ r_1 : r_2 = 9 : 8 \]

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