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Current Question (ID: 19464)

Question:

$\text{Three identical particles } A, B, \text{ and } C \text{ of mass } 100 \text{ kg each are placed in a straight line with } AB = BC = 13 \text{ m.}$ $\text{The gravitational force on a fourth particle } P \text{ of the same mass is } F, \text{ when placed at a distance of } 13 \text{ m from particle } B \text{ on the perpendicular bisector of the line } AC. \text{ The value of } F \text{ will be approximately:}$

Options:

1. $21G$
2. $100G$
3. $59G$
4. $42G$

Solution:

$\text{Hint: Apply Newton's law of gravitation.}$ $\text{Step: Find the value of the force } F$ $\text{The gravitational force between the different particles is shown in the figure below;} \ F_{\text{net}} = \sqrt{2}F_1 + F_2$ $\text{The gravitational force between two particles is given by;} \ F = \frac{GMM}{r^2}$ $\text{The net gravitational force is given by;} \ F = F_2 + \sqrt{2}F_1$ $F = \frac{GMM}{d^2} + \sqrt{2} \frac{GMM}{(\sqrt{2}d)^2}$ $F = \frac{GMM}{d^2} \left(1 + \frac{1}{\sqrt{2}}\right)$ $F = \frac{G \times 10^4}{13^2} \left(1 + \frac{1}{\sqrt{2}}\right)$ $F \approx 100G$ $\text{Hence, option (2) is the correct answer.}$

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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
}