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Current Question (ID: 20287)

Question:

$\text{What will be the most suitable combination of three resistors}$ $A = 2 \, \Omega, \ B = 4 \, \Omega, \ C = 6 \, \Omega \ \text{so that} \ \left( \frac{22}{3} \right) \, \Omega \ \text{is the equivalent resistance of combination?}$

Options:

1. $\text{Parallel combination of} \ A \ \text{and} \ C \ \text{connected in series with} \ B.$
2. $\text{Parallel combination of} \ A \ \text{and} \ B \ \text{connected in series with} \ C.$
3. $\text{Series combination of} \ A \ \text{and} \ C \ \text{connected in parallel with} \ B.$
4. $\text{Series combination of} \ B \ \text{and} \ C \ \text{connected in parallel with} \ A.$

Solution:

$\text{Hint: Recall series and parallel combinations of resistors.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}