Import Question JSON

Current Question (ID: 20318)

Question:

$10$ $\text{resistors each of } 10 \, \Omega \text{ resistance when connected together give}$ $\text{minimum equivalent resistance } R_1 \text{ and maximum equivalent}$ $\text{resistance } R_2 \text{ among various possible combinations.}$ $\text{So, } \frac{R_2}{R_1} \text{ is equal to:}$

Options:

1. $1$
2. $100$
3. $200$
4. $10$

Solution:

$\text{Hint: Use the concept of series and parallel combinations.}$ $R_{\text{min}} = \frac{R}{10} = 1\, \Omega \text{ (when all resistors are placed in parallel)}$ $R_{\text{max}} = 10R = 100\, \Omega \text{ (when all resistors are placed in series)}$ $\Rightarrow \frac{R_{\text{max}}}{R_{\text{min}}} = 100$

Import JSON File

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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
}