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Current Question (ID: 19440)

Question:

$\text{The mass density of a spherical galaxy varies as } \frac{K}{r} \text{ over a large distance } r \text{ from its centre.}$ $\text{In that region, a small star is in a circular orbit of radius } R. \text{ Then the period of revolution, } T \text{ depends on } R \text{ as:}$

Options:

1. $T \propto R$
2. $T^2 \propto \frac{1}{R^3}$
3. $T^2 \propto R$
4. $T^2 \propto R^3$

Solution:

$\text{Hint: } \frac{GMm}{R^2} = m\omega^2 R$ $dm = \rho dv$ $dm = \left( \frac{k}{r} \right) \left( 4\pi r^2 \, dr \right)$ $dm = 4\pi kr \, dr$ $M = \int_0^R dm = \int_0^R 4\pi kr \, dr$ $M = 4\pi k \frac{r^2}{2} \bigg|_0^R$ $M = 2\pi k (R^2 - 0)$ $M = 2\pi k R^2$ $\text{For circular motion gravitational force will provide required centripetal force or}$ $\frac{GMm}{R^2} = \frac{mv^2}{R}$ $G(2\pi k R^2)m = \frac{mv^2}{R} = v = \sqrt{2\pi GkR}$ $T = \frac{2\pi R}{v}$ $T = \frac{2\pi R}{\sqrt{2\pi GkR}} \propto \sqrt{R}$ $\text{or } T^2 \propto R$

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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
}