To answer the question, let's break it down into two parts:
### Part 1: Calculating Linear Expansivity of the Metal Rod
Given:
- Initial length of the rod [tex]\( L_0 = 40 \, \text{cm} \)[/tex]
- Final length of the rod [tex]\( L_1 = 40.05 \, \text{cm} \)[/tex]
- Initial temperature [tex]\( T_0 = 0^\circ \text{C} \)[/tex]
- Final temperature [tex]\( T_1 = 45^\circ \text{C} \)[/tex]
The change in length of the rod is:
[tex]\[ \Delta L = L_1 - L_0 = 40.05 \, \text{cm} - 40 \, \text{cm} = 0.05 \, \text{cm} \][/tex]
The change in temperature is:
[tex]\[ \Delta T = T_1 - T_0 = 45^\circ \text{C} - 0^\circ \text{C} = 45 \, \text{K} \][/tex]
Linear expansivity ([tex]\(\alpha\)[/tex]) is given by the formula:
[tex]\[ \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \][/tex]
Substituting the known values:
[tex]\[ \alpha = \frac{0.05 \, \text{cm}}{40 \, \text{cm} \cdot 45 \, \text{K}} \][/tex]
[tex]\[ \alpha = \frac{0.05}{40 \times 45} \][/tex]
[tex]\[ \alpha = \frac{0.05}{1800} \][/tex]
[tex]\[ \alpha = \frac{5 \times 10^{-2}}{1.8 \times 10^3} \][/tex]
[tex]\[ \alpha = \frac{5}{1800} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{360} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{3.6 \times 10^2} \times 10^{-2} \][/tex]
[tex]\[ \alpha = \frac{1}{3.6} \times 10^{-5} \][/tex]
[tex]\[ \alpha \approx 2.78 \times 10^{-5} \, \text{K}^{-1} \][/tex]
### Part 2: Calculating P.d Across the Terminal
Given:
- Electromotive force (E.M.F): [tex]\( \mathcal{E} = 12 \, \text{V} \)[/tex]
- Internal resistance: [tex]\( r = 2 \, \Omega \)[/tex]
- External resistance: [tex]\( R = 8 \, \Omega \)[/tex]
First, calculate the total resistance in the circuit:
[tex]\[ R_{\text{total}} = r + R = 2 \Omega + 8 \Omega = 10 \Omega \][/tex]
Next, use Ohm's Law to calculate the current ([tex]\(I\)[/tex]) flowing through the circuit:
[tex]\[ I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12 \, \text{V}}{10 \, \Omega} = 1.2 \, \text{A} \][/tex]
Finally, calculate the potential difference (P.d) across the terminal using the formula:
[tex]\[ V_{\text{terminal}} = \mathcal{E} - I \cdot r = 12 \, \text{V} - 1.2 \, \text{A} \times 2 \, \Omega = 12 \, \text{V} - 2.4 \, \text{V} = 9.6 \, \text{V} \][/tex]
Thus, the P.d across the terminal is:
[tex]\[ \boxed{9.6 \, \text{V}} \][/tex]
The correct answer for the second problem is: B. [tex]\( 9.6 \, \text{V} \)[/tex]
