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Current Question (ID: 19493)

Question:

$\text{The mass density of a planet of radius } R \text{ varies with the distance } r$ $\text{from its centre as } \rho(r) = \rho_0 \left(1 - \frac{r^2}{R^2}\right)$ $\text{Then the gravitational field is maximum at:}$

Options:

1. $r = \sqrt{\frac{3}{4}} R$
2. $r = \sqrt{\frac{5}{9}} R$
3. $r = R$
4. $r = \frac{1}{\sqrt{3}} R$

Solution:

$\text{Hint: Total mass (M) of the planet } = \int \rho(r) 4 \pi r^2 dr.$ $E = 4 \pi r^2 = \int \rho_0 4 \pi r^2 dr$ $E r^2 = 4 \pi G \int_0^r \rho_0 \left(1 - \frac{r^2}{R^2}\right) r^2 dx$ $E = 4 \pi G \rho_0 \left(\frac{r^3}{3} - \frac{r^5}{5R^2}\right)$ $\frac{dE}{dr} = 0 \therefore r = \ldots$

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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
}