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Current Question (ID: 19910)

Question:

$\text{Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance } (R/2) \text{ from the earth's center, where } 'R' \text{ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:}$

Options:

1. $\frac{2\pi R}{g}$
2. $\frac{g}{2\pi R}$
3. $\frac{1}{2\pi} \sqrt{\frac{g}{R}}$
4. $2\pi \sqrt{\frac{R}{g}}$

Solution:

$\text{Hint: Express acceleration as } \omega^2 x$ $\text{Force along the tunnel}$ $F = -\left(\frac{GMmr}{R^3}\right) \cos \theta$ $F = -\frac{gm}{R} x \left(\frac{GM}{R^2} = g, \; r \cos \theta = x\right)$ $a = -\frac{g}{R} x$ $\omega^2 = \frac{g}{R} \; T = 2\pi \sqrt{\frac{R}{g}}$

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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
}