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Current Question (ID: 20436)

Question:

$\text{As shown in the figure, a metallic rod of linear density } 0.45 \text{ kg m}^{-1} \text{ is lying horizontally on a smooth incline plane which makes an angle of } 45^\circ \text{ with the horizontal.}$ $\text{The minimum current flowing in the rod required to keep it stationary, when } 0.15 \text{ T magnetic field is acting on it in the vertically upward direction, will be:}$ $\{\text{Use } g = 10 \text{ m/s}^2 \}$

Options:

1. $30 \text{ A}$
2. $15 \text{ A}$
3. $10 \text{ A}$
4. $3 \text{ A}$

Solution:

$\text{The force due to gravity on the rod is } F_g = \lambda g \sin \theta = 0.45 \times 10 \times \sin 45^\circ = 3.18 \text{ N}$ $\text{The magnetic force required to balance this is } F_B = BIL \sin \theta$ $\text{Equating the forces: } BIL \sin \theta = \lambda g \sin \theta$ $I = \frac{\lambda g}{B} = \frac{0.45 \times 10}{0.15} = 30 \text{ A}$

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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
}