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Current Question (ID: 19461)

Question:
$\text{Which of the following graphs correctly represents the variation of}$ $\text{acceleration due to gravity } g \text{ with distance } r \text{ from the centre of the}$ $\text{Earth? Assume } R \text{ is the radius of the Earth.}$
Options:
  • 1. $\text{Graph 1}$
  • 2. $\text{Graph 2}$
  • 3. $\text{Graph 3}$
  • 4. $\text{Graph 4}$
Solution:
$\text{Hint: } g_h = g \left(1 - \frac{2h}{R}\right), \ g_d = g \left(1 - \frac{d}{R}\right)$ $\text{Step 1: Find the variation of acceleration due to gravity with}$ $\text{altitude } (r > R).$ $g' = \frac{gR^2}{r^2}$ $\Rightarrow g' \propto \frac{1}{r^2}$ $\text{Step 2: Find the variation of acceleration due to gravity with depth}$ $(r < R).$ $g' = \frac{gr}{R}$ $\Rightarrow g' \propto r$ $\text{Step 3: Identify the correct graph.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}