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Current Question (ID: 16993)

Question:

$\text{A rod } AB \text{ of length } l \text{ is moving with constant speed } v \text{ in a uniform magnetic field on a conducting } U\text{-shaped wire as shown.}$ $\text{If the rate of loss of heat energy across resistance } R \text{ is } Q, \text{ then the force needed parallel to velocity to keep rod moving with constant speed } v \text{ is:}$

Options:

1. $Qv$
2. $\frac{Q}{v}$
3. $\frac{Q^2}{v}$
4. $Q^2v$

Solution:

$\text{The power dissipated across the resistor is given by } P = \frac{Q}{t}.$ $\text{The induced emf } \varepsilon = Bvl.$ $\text{The current } I = \frac{\varepsilon}{R} = \frac{Bvl}{R}.$ $\text{The power } P = I^2R = \left(\frac{Bvl}{R}\right)^2 R = \frac{B^2v^2l^2}{R}.$ $\text{Equating the power } P \text{ to } Q, \text{ we get } \frac{B^2v^2l^2}{R} = Q.$ $\text{The force } F = BIl = B \left(\frac{Bvl}{R}\right) l = \frac{B^2v l^2}{R}.$ $\text{Substituting } \frac{B^2v^2l^2}{R} = Q, \text{ we get } F = \frac{Q}{v}.$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}