Q154 / 180 · COMEDK 2024 (Morning Shift) · Physics
The current through a conductor is '$\mathrm{a}$' when the temperature is $0^{\circ} \mathrm{C}$. It is '$\mathrm{b}$' when the temperature is $100^{\circ} \mathrm{C}$. The current through the conductor at $220^{\circ} \mathrm{C}$ is
- A$\dfrac{11 a b}{5 a-6 b}$
- B$\dfrac{5 a b}{6 a-11 b}$
- Correct answerC$\dfrac{5 a b}{11 a-6 b}$
- D$\dfrac{5 a b}{11 b-6 a}$
Solution
The resistance of a conductor at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$, where $R_0$ is the resistance at $0^{\circ} \text{C}$ and $\alpha$ is the temperature coefficient of resistance.
Given the current $I = \dfrac{V}{R}$, where $V$ is the constant applied voltage, we have $I_T = \dfrac{V}{R_0(1 + \alpha T)} = \dfrac{I_0}{1 + \alpha T}$.
At $T = 0^{\circ} \text{C}$, $I_0 = a$. Thus, $I_T = \dfrac{a}{1 + \alpha T}$.
At $T = 100^{\circ} \text{C}$, $I = b$. Substituting this into the equation: $b = \dfrac{a}{1 + 100\alpha} \Rightarrow 1 + 100\alpha = \dfrac{a}{b} \Rightarrow 100\alpha = \dfrac{a}{b} - 1 = \dfrac{a - b}{b} \Rightarrow \alpha = \dfrac{a - b}{100b}$.
We need the current at $T = 220^{\circ} \text{C}$. Let this be $I_{220}$.
$I_{220} = \dfrac{a}{1 + 220\alpha} = \dfrac{a}{1 + 220 \times \left( \dfrac{a - b}{100b} \right)} = \dfrac{a}{1 + \dfrac{22}{10} \times \dfrac{a - b}{b}} = \dfrac{a}{1 + \dfrac{11(a - b)}{5b}}$.
$I_{220} = \dfrac{a}{\dfrac{5b + 11a - 11b}{5b}} = \dfrac{5ab}{11a - 6b}$.
Answer: $\dfrac{5 a b}{11 a-6 b}$
Given the current $I = \dfrac{V}{R}$, where $V$ is the constant applied voltage, we have $I_T = \dfrac{V}{R_0(1 + \alpha T)} = \dfrac{I_0}{1 + \alpha T}$.
At $T = 0^{\circ} \text{C}$, $I_0 = a$. Thus, $I_T = \dfrac{a}{1 + \alpha T}$.
At $T = 100^{\circ} \text{C}$, $I = b$. Substituting this into the equation: $b = \dfrac{a}{1 + 100\alpha} \Rightarrow 1 + 100\alpha = \dfrac{a}{b} \Rightarrow 100\alpha = \dfrac{a}{b} - 1 = \dfrac{a - b}{b} \Rightarrow \alpha = \dfrac{a - b}{100b}$.
We need the current at $T = 220^{\circ} \text{C}$. Let this be $I_{220}$.
$I_{220} = \dfrac{a}{1 + 220\alpha} = \dfrac{a}{1 + 220 \times \left( \dfrac{a - b}{100b} \right)} = \dfrac{a}{1 + \dfrac{22}{10} \times \dfrac{a - b}{b}} = \dfrac{a}{1 + \dfrac{11(a - b)}{5b}}$.
$I_{220} = \dfrac{a}{\dfrac{5b + 11a - 11b}{5b}} = \dfrac{5ab}{11a - 6b}$.
Answer: $\dfrac{5 a b}{11 a-6 b}$
Hence, the correct answer is option: 'C'.