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Current Question (ID: 8767)

Question:

\text{Two resistors } R_1 = (3.0 \pm 0.3) \, \Omega \text{ and } R_2 = (5.0 \pm 0.1) \, \Omega \text{ are connected in parallel.} \text{The equivalent resistance, } R_{\text{eq}}\text{, will be:} \text{Hint: } \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}

Options:

1. $1.9 \pm 0.07 \Omega$
2. $1.9 \pm 0.1 \Omega$
3. $2.9 \pm 0.2 \Omega$
4. $2.9 \pm 0.3 \Omega$

Solution:

\text{Hint: } \frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \text{Step 1: Find the mean equivalent resistance.} \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} R_{\text{eq}} = \frac{3 \times 5}{3 + 5} \approx 1.9 \, \Omega \text{Step 2: Find the mean absolute error in equivalent resistance.} \frac{\Delta R_{\text{eq}}}{R_{\text{eq}}^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \Delta R_{\text{eq}} = (1.9)^2 \left[ \frac{0.3}{3^2} + \frac{0.1}{5^2} \right] \Delta R_{\text{eq}} = 0.1 \, \Omega \text{Step 3: Report the equivalent resistance.} \text{Result} = R_{\text{eq}} \pm \Delta R_{\text{eq}} \text{Result} = (1.9 \pm 0.1) \, \Omega \text{Hence, option (2) 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
}