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Current Question (ID: 16730)

Question:

$\text{The net resistance of the circuit between } A \text{ and } B \text{ is:}$

Options:

1. $\frac{8}{3} \, \Omega$
2. $\frac{14}{3} \, \Omega$
3. $\frac{16}{3} \, \Omega$
4. $\frac{22}{3} \, \Omega$

Solution:

$\text{Hint: Apply the concept of Wheatstone Bridge.}$ $\text{Step: Find the net resistance of the circuit between } A \text{ and } B$ $\text{Since it is a balanced Wheatstone bridge, the equivalent resistance between } A \text{ and } B \text{ is given by:}$ $\text{When } 3 \, \Omega \text{ and } 4 \, \Omega \text{ are in series, the total resistance:}$ $\Rightarrow R_1 = (3 + 4) \, \Omega = 7 \, \Omega$ $\text{When } 6 \, \Omega \text{ and } 8 \, \Omega \text{ are in series, the total resistance:}$ $\Rightarrow R_2 = (6 + 8) \, \Omega = 14 \, \Omega$ $\text{So total resistance when } R_1 \text{ and } R_2 \text{ are in parallel:}$ $\Rightarrow R = \frac{R_1 \times R_2}{R_1 + R_2}$ $\text{Substitute the known values we get:}$ $\Rightarrow R = \frac{14 \times 7}{21} = \frac{14}{3} \, \Omega$ $\text{Hence, option (2) is the correct answer.}$

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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
}