Two wires of the same material have length 6cm and 10cm and radii 0.5 mm and 1.5 mm respectively. They are connected in series across a battery of 16 V. The p.d. across the shorter wire is

Options

A. 5V
B. 13.5 V
C. 27 V
D. 10 V

Answer

To solve the problem step by step, we will calculate the potential difference (p.d.) across the shorter wire (length 6 cm) when two wires are connected in series across a 16 V battery.

### Step 1: Identify given values
- Length of wire 1 (shorter wire, \( l_1 \)) = 6 cm = 0.06 m
- Length of wire 2 (longer wire, \( l_2 \)) = 10 cm = 0.10 m
- Radius of wire 1 (\( r_1 \)) = 0.5 mm = 0.0005 m
- Radius of wire 2 (\( r_2 \)) = 1.5 mm = 0.0015 m
- Voltage of the battery (\( V \)) = 16 V

### Step 2: Calculate the cross-sectional areas of the wires
The cross-sectional area \( A \) of a wire is given by the formula:
\[ A = \pi r^2 \]
- For wire 1:
\[ A_1 = \pi (0.0005)^2 = \pi \times 0.00000025 \approx 7.85 \times 10^{-7} \, \text{m}^2 \]
- For wire 2:
\[ A_2 = \pi (0.0015)^2 = \pi \times 0.00000225 \approx 7.07 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the resistances of the wires
The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho l}{A} \]
Since both wires are made of the same material, they have the same resistivity \( \rho \).

- Resistance of wire 1 (\( R_1 \)):
\[ R_1 = \frac{\rho l_1}{A_1} = \frac{\rho \times 0.06}{7.85 \times 10^{-7}} \]
- Resistance of wire 2 (\( R_2 \)):
\[ R_2 = \frac{\rho l_2}{A_2} = \frac{\rho \times 0.10}{7.07 \times 10^{-6}} \]

### Step 4: Find the ratio of the resistances
From the equations for \( R_1 \) and \( R_2 \):
\[ \frac{R_1}{R_2} = \frac{\frac{\rho \times 0.06}{A_1}}{\frac{\rho \times 0.10}{A_2}} = \frac{0.06 \cdot A_2}{0.10 \cdot A_1} \]
Substituting the values of \( A_1 \) and \( A_2 \):
\[ \frac{R_1}{R_2} = \frac{0.06 \cdot 7.07 \times 10^{-6}}{0.10 \cdot 7.85 \times 10^{-7}} = \frac{0.06 \cdot 7.07}{0.10 \cdot 7.85} \approx 0.054 \]

### Step 5: Use the voltage division rule
Since the wires are in series, the voltage across each wire can be calculated using the voltage division rule:
\[ \frac{V_1}{V_2} = \frac{R_1}{R_2} \]
Let \( V_1 \) be the potential difference across the shorter wire and \( V_2 \) across the longer wire. Thus:
\[ V_1 = \frac{R_1}{R_1 + R_2} \cdot V \]
Substituting \( V = 16 \, \text{V} \):
\[ V_1 = \frac{R_1}{R_1 + R_2} \cdot 16 \]

### Step 6: Calculate \( V_1 \)
Using the ratio we found:
\[ V_1 + V_2 = 16 \implies V_2 = 16 - V_1 \]
Substituting into the voltage ratio:
\[ V_1 = \frac{R_1}{R_1 + R_2} \cdot 16 \]
Substituting the values of \( R_1 \) and \( R_2 \) based on their ratios gives:
\[ V_1 = \frac{0.054}{0.054 + 1} \cdot 16 \]
Calculating gives:
\[ V_1 \approx 2.5 \, \text{V} \]

### Final Answer
The potential difference across the shorter wire is approximately **2.5 V**.