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Current Question (ID: 16672)

Question:

$\text{The charge flowing through a resistance } R \text{ varies with time } t \text{ as}$ $Q = at - bt^2, \text{ where } a \text{ and } b \text{ are positive constants. The total heat produced in } R \text{ is:}$

Options:

1. $\frac{a^3 R}{3b}$
2. $\frac{a^3 R}{2b}$
3. $\frac{a^3 R}{b}$
4. $\frac{a^3 R}{6b}$

Solution:

$\text{Hint: } H = \int_0^t I^2 R dt$ $\text{Step: Find the total heat produced in } R$ $\text{The charge } Q \text{ is given as:}$ $Q = at - bt^2 \quad \cdots (1)$ $I = \frac{dQ}{dt} = a - 2bt$ $\text{The maximum value of } t \text{ for which the current exists can be calculated as:}$ $\Rightarrow a - 2bt = 0$ $\therefore t = \frac{a}{2b} \quad \cdots (2)$ $\text{The total heat produced in } R \text{ from } t = \frac{a}{2b} \text{ will be equal to:}$ $H = \int_0^{t} I^2 R dt$ $H = \int_0^{a/2b} (a - 2bt)^2 R \cdot dt$ $H = \int_0^{a/2b} (a^2 + 4b^2 t^2 - 4abt) R dt$ $H = \left( a^2 t + 4b^2 \frac{t^3}{3} - \frac{4abt^2}{2} \times 0 \right) R$ $\text{Solving the above equation we get,}$ $\Rightarrow H = \frac{a^3 R}{6b}$ $\text{Hence, option (4) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}