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Current Question (ID: 19450)

Question:

$\text{Four identical particles of equal masses } 1 \text{ kg made to move along the circumference of a circle of radius } 1 \text{ m under the action of their own mutual gravitational attraction. The speed of each particle will be:}$

Options:

1. $\sqrt{\frac{G}{2} \left(1 + 2\sqrt{2}\right)}$
2. $\sqrt{G \left(1 + 2\sqrt{2}\right)}$
3. $\sqrt{\frac{G}{2} \left(2\sqrt{2} - 1\right)}$
4. $\sqrt{\frac{(1 + 2\sqrt{2})G}{2}}$

Solution:

$\text{Hint: } F_{\text{net}} = \frac{mv^2}{r}$ $F_1 = \frac{Gmm}{2(R)^2} = \frac{Gm^2}{4R^2}$ $F_2 = \frac{Gmm}{(\sqrt{2}R)^2} = \frac{Gm^2}{2R^2}$ $F_3 = \frac{Gmm}{(\sqrt{2}R)^2} = \frac{Gm^2}{2R^2}$ $\Rightarrow F_{\text{net}} = F_1 + F_2 \cos 45^\circ + F_3 \cos 45^\circ$ $= \frac{Gm^2}{4R^2} + \frac{Gm^2}{2R^2} \frac{1}{\sqrt{2}} + \frac{Gm^2}{2R^2} \frac{1}{\sqrt{2}}$ $= \frac{Gm^2}{R^2} \left(\frac{1}{4} + \frac{1}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}\right)$ $= \frac{Gm^2}{R^2} \left(\frac{1}{4} + \frac{1}{\sqrt{2}}\right) = \frac{Gm^2}{4R^2} \left(1 + 2\sqrt{2}\right)$ $F_{\text{net}} = \frac{Gm^2}{4R^2} \left(1 + 2\sqrt{2}\right) = \frac{mv^2}{R}$ $\Rightarrow v = \sqrt{\frac{G \left(1 + 2\sqrt{2}\right)}{2}}$

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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
}