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Current Question (ID: 19943)

Question:

$\text{A pendulum bob is released from angle } \theta \text{ with the vertical as shown in the figure. If its acceleration at maximum amplitude is the same as at mean position, then find } \theta.$

Options:

1. $\tan^{-1}(\sqrt{2})$
2. $2 \tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$
3. $2 \tan^{-1}\left(\frac{1}{2}\right)$
4. $\tan^{-1}(2)$

Solution:

$\text{Hint: } v^2 = 2gl(1 - \cos \theta)$ $\text{At max amplitude } a_1 = g \sin \theta$ $\text{at mean position } a_2 = \frac{v^2}{l}$ $\therefore v^2 = 2gl(1 - \cos \theta)$ $\therefore a_2 = 2g(1 - \cos \theta)$ $g \sin \theta = 2g(1 - \cos \theta)$ $2g \sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2g \times 2 \sin^2 \frac{\theta}{2}$ $\cos \frac{\theta}{2} = 2 \sin \frac{\theta}{2}$ $\tan \frac{\theta}{2} = \frac{1}{2}$ $\theta = 2 \tan^{-1}\left(\frac{1}{2}\right)$

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
}