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Current Question (ID: 19117)

Question:

$\text{For a particle in a uniform circular motion, the acceleration } \vec{a} \text{ at any point } P(R, \theta) \text{ on the circular path of radius } R \text{ is: (when } \theta \text{ is measured from the positive x-axis and } v \text{ is uniform speed):}$

Options:

1. $-\frac{v^2}{R} \sin \theta \hat{i} + \frac{v^2}{R} \cos \theta \hat{j}$
2. $-\frac{v^2}{R} \cos \theta \hat{i} + \frac{v^2}{R} \sin \theta \hat{j}$
3. $-\frac{v^2}{R} \cos \theta \hat{i} - \frac{v^2}{R} \sin \theta \hat{j}$
4. $-\frac{v^2}{R} \hat{i} + \frac{v^2}{R} \hat{j}$

Solution:

$\text{In uniform circular motion, the acceleration is directed towards the center of the circle.}$ $\text{The radial acceleration is given by } \vec{a} = -\frac{v^2}{R} \hat{r}.$ $\text{In terms of } \theta, \text{ the unit vector } \hat{r} \text{ is } \cos \theta \hat{i} + \sin \theta \hat{j}.$ $\text{Thus, } \vec{a} = -\frac{v^2}{R} (\cos \theta \hat{i} + \sin \theta \hat{j}) = -\frac{v^2}{R} \cos \theta \hat{i} - \frac{v^2}{R} \sin \theta \hat{j}.$

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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
}