Import Question JSON

Current Question (ID: 9555)

Question:

$\text{A boy runs on a circular track of radius } R \text{ (in km) with a speed of } \frac{\pi R}{2} \text{ km/h in the clockwise direction for 3 h and then with } \pi R \text{ km/h in the anticlockwise direction for 1 h. The magnitude of his displacement will be:}$

Options:

1. $\frac{\pi R}{2}$
2. $\frac{R}{\sqrt{2}}$
3. $\frac{3\pi R}{2}$
4. $\sqrt{2R}$

Solution:

\text{Hint: Find the final position of the boy.} \text{Step 1: Find the final position for the clockwise motion.} \text{For the clockwise direction, the distance covered by the particle is:} \text{Distance} = \text{speed} \times \text{time} = \frac{\pi R}{2} \times 3 = \frac{3\pi R}{2} \text{So, the particle covers three-quarters of the circle in the clockwise direction and reaches point B starting from point A.} \text{Step 2: Find the final position for anti-clockwise motion.} \text{In the anti-clockwise direction, the distance covered by the particle is:} \text{Distance} = \pi R \times 1 = \pi R \text{So, the particle covers two quarters in the anti-clockwise direction and reaches point C starting from point B.} \text{So, the initial position of the particle is A and the final position of the particle is C.} \text{From the Pythagoras theorem, the displacement is given by:} |\vec{r}| = \sqrt{2R} \text{Hence, option (4) is the correct answer.}

Import JSON File

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
}