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Current Question (ID: 19454)

Question:

$\text{A solid sphere of radius } R \text{ gravitationally attracts a particle placed at } 3R \text{ form its centre with a force } F_1. \text{ Now a spherical cavity of radius } \frac{R}{2} \text{ is made in the sphere (as shown in figure) and the force becomes } F_2. \text{ The value of } F_1 : F_2 \text{ is:}$

Options:

1. $25 : 36$
2. $36 : 25$
3. $50 : 41$
4. $41 : 50$

Solution:

$\text{Hint: } F = \frac{Gm_1m_2}{r^2}$ $\text{Let initial mass of sphere is } m'. \text{ Hence mass of removed portion will be } \frac{m'}{8}$ $F_1 = m \cdot E. = \frac{m \cdot Gm'}{9R^2}$ $F_2 = m \left[ \frac{G \cdot m'}{(3R)^2} - \frac{G \cdot m'/8}{(5R/2)^2} \right]$ $= \frac{Gm'}{9R^2} - \frac{Gm' \cdot 4}{8 \times 25}$ $= \left( \frac{1}{9} - \frac{1}{50} \right) \frac{Gm'}{R^2}$ $F_2 = \frac{41}{50 \times 9} \cdot \frac{Gm'}{R^2}$ $\frac{F_1}{F_2} = \frac{1}{9} \times \frac{50 \times 9}{41} = \frac{50}{41}$

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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
}