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Current Question (ID: 16992)

Question:
$\text{A rod having length } l \text{ and resistance } R_0 \text{ is moving with a speed } v \text{ as shown in the figure. The current through the rod is:}$
Options:
  • 1. $\frac{Blv}{\frac{R_1 R_2}{R_1 + R_2} + R_0}$
  • 2. $\frac{Blv}{\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_0}\right)^2}$
  • 3. $\frac{Blv}{R_1 + R_2 + R_0}$
  • 4. $\frac{Blv}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_0}}$
Solution:
$\text{Hint: Recall the concept of mutual induction.}$ $\text{Step 1: Find the induced emf in the rod.}$ $\text{The induced emf in the rod is given by:}$ $\varepsilon = Blv$ $\text{Step 2: Find the current through the rod.}$ $R' = \frac{R_1 R_2}{R_1 + R_2}$ $i = \frac{\varepsilon}{\frac{R_1 R_2}{R_1 + R_2} + R_0}$ $\Rightarrow i = \frac{Blv}{\frac{R_1 R_2}{R_1 + R_2} + R_0}$ $\text{Hence, option (1) is the correct answer.}$

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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
}