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Current Question (ID: 19513)

Question:

$\text{If } W \text{ is the weight on the surface of the Earth, then the weight of the same body at a height of } \frac{R_e}{4} \text{ above the surface of Earth is equal to: } (R_e = \text{radius of Earth})$

Options:

1. $\frac{4}{5}W$
2. $\frac{16}{25}W$
3. $\frac{25}{16}W$
4. $\frac{5}{4}W$

Solution:

$\text{Hint: Applying Newton's law of gravitation.}$ $\text{Step: Find the weight of the body at height } \frac{R_e}{4}. \text{ The acceleration due to gravity at height } h \text{ is given by:}$ $g' = \frac{g}{\left(1 + \frac{h}{R_E}\right)^2}$ $= \frac{gR_E^2}{\left(\frac{5R_E}{4}\right)^2} = \frac{16g}{25}$ $\text{The weight of the body at height } \frac{R_e}{4} \text{ is } \frac{16W}{25}. \text{ Hence, option (2) is the correct answer.}$

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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
}