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Current Question (ID: 19467)

Question:
$\text{If the weight of an object at the Earth's surface is } 18 \text{ N, what will be its weight when taken to a height of } 3200 \text{ km above the surface?}$ $\text{(Given: radius of Earth } = 6400 \text{ km)}$
Options:
  • 1. $9 \text{ N}$
  • 2. $8 \text{ N}$
  • 3. $20 \text{ N}$
  • 4. $18 \text{ N}$
Solution:
$\text{Hint: } g = \frac{GM}{r^2}$ $\text{Step: Find the weight of the given object.}$ $\text{As we know that;}$ $g = \frac{GM}{r^2}$ $g_{\text{new}} = \frac{GM}{\left(R + \frac{R}{2}\right)^2} = \frac{4}{9} \times \frac{GM}{R^2} = \frac{4}{9} g_{\text{surface}}$ $\text{New weight } = \frac{4}{9} \times 18 \text{ N} = 8 \text{ N}$ $\text{Hence, option (2) is the correct answer.}$

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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
}