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Current Question (ID: 19457)

Question:

$\text{Determine the gravitational force of attraction between a ring and a sphere as shown in the figure, where the plane of the ring is perpendicular to the line joining their centres.}$ $\text{Given that the distance between the centres of the ring (of mass } m \text{) and the sphere (of mass } M \text{) is } \sqrt{8}R, \text{ and both have the same radius } R.$

Options:

1. $\frac{\sqrt{8}}{9} \cdot \frac{GmM}{R}$
2. $\frac{2\sqrt{2}}{3} \cdot \frac{GMm}{R^2}$
3. $\frac{1}{3\sqrt{8}} \cdot \frac{GMm}{R^2}$
4. $\frac{\sqrt{8}}{27} \cdot \frac{GmM}{R^2}$

Solution:

$\text{Hint: Consider the shell as a point mass, placed at its centre.}$ $\text{Step: Find the gravitational force of attraction between the ring and the sphere.}$ $\text{Consider the sphere as a point mass placed at the centre of the sphere.}$ $\text{The ring and point mass are shown in the figure below;}$$\text{The gravitational field due to the ring at a distance } x \text{ is given by:}$ $E = -\frac{Gmx}{(R^2 + x^2)^{\frac{3}{2}}}$ $\text{The gravitational force of attraction between the ring and the sphere is given by:}$ $F = \frac{GMmx}{(R^2 + x^2)^{\frac{3}{2}}}$ $\Rightarrow F = \frac{GMm\sqrt{8}R}{(R^2 + (\sqrt{8}R)^2)^{\frac{3}{2}}} = \frac{GMm\sqrt{8}R}{(R^2 + 8R^2)^{\frac{3}{2}}}$ $\Rightarrow F = \frac{GMm\sqrt{8}R}{(9R^2)^{\frac{3}{2}}}$ $\Rightarrow F = \frac{GMm\sqrt{8}R}{27R^3} = \frac{\sqrt{8}GMm}{27R^2}$ $\Rightarrow F = \frac{\sqrt{8}}{27} \cdot \frac{GMm}{R^2}$ $\text{Hence, option (4) is the correct answer.}$

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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
}