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Current Question (ID: 20228)

Question:

$\text{A current through a wire depends on time as } \dot{i} = \alpha_0 t + \beta t^2 \text{ where } \alpha_0 = 20 \text{ A/s and } \beta = 8 \text{ As}^{-2}. \text{ The charge crossed through a cross-section of wire in } 15 \text{ s is:}$

Options:

1. $2250 \text{ C}$
2. $11250 \text{ C}$
3. $2100 \text{ C}$
4. $260 \text{ C}$

Solution:

$\text{Hint: } i = \frac{dq}{dt}$ $\text{Step 1: Use formula for current}$ $i = \frac{dq}{dt}$ $\text{Step 2: Integrate on both sides for } t = 0 \text{ to } t = 15 \text{ s}$ $\int_0^q dq = \int_0^{15} [20t + 8t^2] \, dt$ $q = \left[ 10t^2 + \frac{8}{3} t^3 \right]_0^{15}$ $= 10(15)^2 + \frac{8}{3}(15)^3$ $= 11250 \text{ C}$

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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
}