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Current Question (ID: 16983)

Question:
$\text{A wire } cd \text{ of length } l \text{ and mass } m \text{ is sliding without friction on conducting rails } ax \text{ and } by \text{ as shown.}$ $\text{The vertical rails are connected to each other with a resistance } R \text{ between } a \text{ and } b.$ $\text{A uniform magnetic field } B \text{ is applied perpendicular to the plane } abcd \text{ such that } cd \text{ moves with a constant velocity of:}$
Options:
  • 1. $\frac{mgR}{Bl}$
  • 2. $\frac{mgR}{B^2l^2}$
  • 3. $\frac{mgR}{B^3l^3}$
  • 4. $\frac{mgR}{B^2l}$
Solution:
$(2) \text{ Due to magnetic field, wire will experience an upward force }$ $F = Bil = B \left( \frac{Bvl}{R} \right) l$ $\Rightarrow F = \frac{B^2vl^2}{R}$ $\text{If wire slides down with constant velocity, then -}$ $F = mg \Rightarrow \frac{B^2vl^2}{R} = mg \Rightarrow v = \frac{mgR}{B^2l^2}$

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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
}