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Current Question (ID: 19452)

Question:
$\text{A body weighs } 49 \text{ N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? (use } g = \frac{GM}{R^2} = 9.8 \text{ ms}^{-2} \text{ radius of earth, } R = 6400 \text{ km.})$
Options:
  • 1. $49 \text{ N}$
  • 2. $48.83 \text{ N}$
  • 3. $49.83 \text{ N}$
  • 4. $49.17 \text{ N}$
Solution:
$\text{Hint: The weight of an object gradually decreases from poles to equator.}$ $\text{Weight of pole } = mg = 49 \text{ N}$ $\text{At equator due to rotation } = g_e = g - R\omega^2$ $\text{so } W = mg_e = m(g - R\omega^2)$ $\therefore \text{ } W_P > W_e \quad W_P = 49 \text{ N}$ $\text{So, } W_e = 48.83 \text{ N. } \quad W_e < 49 \text{ N}$

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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
}