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Current Question (ID: 16995)

Question:

$\text{A copper rod of mass } m \text{ slides under gravity on two smooth parallel rails } l \text{ distance apart and set at an angle } \theta \text{ to the horizontal as shown in figure.}$ $\text{At the bottom, the rails are joined by a resistance } R. \text{ There is a uniform magnetic field perpendicular to the plane of the rails.}$ $\text{The terminal velocity of the rod is:}$

Options:

1. $\frac{mgR \cos \theta}{B^2 l^2}$
2. $\frac{mgR \sin \theta}{B^2 l^2}$
3. $\frac{mgR \tan \theta}{B^2 l^2}$
4. $\frac{mgR \cot \theta}{B^2 l^2}$

Solution:

$\text{Hint: } F_m = mg \sin \theta \text{ (for terminal velocity).}$ $\text{Step: Find the terminal velocity of the rod.}$ $\text{The induced current is given by:}$ $\varepsilon = Blv$ $I = \frac{\varepsilon}{R} = \frac{Blv}{R}$ $\text{The magnetic force on the current-carrying conductor is given by:}$ $\vec{F}_m = i(\vec{l} \times \vec{B})$ $\text{Balance the forces in equilibrium condition we get:}$ $Mg \sin \theta = ilB$ $Mg \sin \theta = \frac{(Blv)lB}{R}$ $\Rightarrow v = \frac{MgR \sin \theta}{B^2 l^2}$ $\text{Hence, option (2) is the correct answer.}$

Import JSON File

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
}