Import Question JSON

Current Question (ID: 19108)

Question:

$\text{A particle moving in a uniform circular motion of radius 1 m has velocity } 3\hat{j} \text{ m/s at point } B. \text{ What are the velocity } (\vec{v}) \text{ and acceleration } (\vec{a}) \text{ at diametrically opposite point } A?$

Options:

1. $\vec{v}_A = 3\hat{j} \text{ m/s}; \; \vec{a}_A = -9\hat{i} \text{ m/s}^2$
2. $\vec{v}_A = -3\hat{j} \text{ m/s}; \; \vec{a}_A = 9\hat{i} \text{ m/s}^2$
3. $\vec{v}_A = -3\hat{j} \text{ m/s}; \; \vec{a}_A = 9\hat{j} \text{ m/s}^2$
4. $\vec{v}_A = 3\hat{i} \text{ m/s}; \; \vec{a}_A = 9\hat{j} \text{ m/s}^2$

Solution:

$\text{Hint: } a = \frac{v^2}{r}$ $\text{Step 1: Analyse the motion.}$ $\text{Step 2: Find the velocity at point } A.$ $\text{The velocity at point } B \text{ is given, } \vec{v}_B = 3\hat{j} \text{ m/s}$ $\text{The velocity of the particle at } A \text{ is; } \vec{v}_A = -3\hat{j} \text{ m/s}$ $\text{Step 3: Find the acceleration at point } A.$ $\text{The acceleration at this point } B \text{ is given by; } \vec{a}_B = \frac{v_B^2}{R} = \frac{3^2}{1} = 9\hat{i} \text{ m/s}^2$ $\text{The acceleration at the point } A \text{ is } \frac{v_A^2}{R}$ $\vec{a}_A = 9\hat{i} \text{ m/s}^2.$ $\text{Therefore, the velocity and acceleration at the point } A \text{ is } \vec{v}_A = -3\hat{j} \text{ m/s}; \; \vec{a}_A = 9\hat{i} \text{ m/s}^2$ $\text{Hence, option (2) is the correct answer.}$

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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
}