Import Question JSON

Current Question (ID: 19486)

Question:

$\text{If the Earth has no rotational motion, the weight of a person on the equator is } W. \text{ Determine the speed with which the Earth would have to rotate about its axis so that the person at the equator will weigh } \frac{3}{4} W.$ $\text{(The radius of the Earth is } 6400 \text{ km and } g = 10 \text{ m/s}^2)$

Options:

1. $0.28 \times 10^{-3} \text{ rad/s}$
2. $1.1 \times 10^{-3} \text{ rad/s}$
3. $0.83 \times 10^{-3} \text{ rad/s}$
4. $0.63 \times 10^{-3} \text{ rad/s}$

Solution:

$\text{Hint: The acceleration due to gravity at the equator, } g_e = g - \omega^2 R \cos^2 \theta$

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
}