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Current Question (ID: 9653)

Question:

$\text{In a uniform circular motion, if the speed of the particle is 2 m/s and radius of the circle is 2 m, then the values of centripetal and tangential acceleration are, respectively:}$

Options:

1. $2 \text{ m/s}^2, 2 \text{ m/s}^2$
2. $2 \text{ m/s}^2, 1 \text{ m/s}^2$
3. $0, 2 \text{ m/s}^2$
4. $2 \text{ m/s}^2, 0$

Solution:

$\text{Hint: In a uniform circular motion, tangential acceleration is zero.}$ $\text{Step 1: Draw the diagram to visualise the direction of acceleration}$ $\text{The diagram shows a circle with velocity vector } v \text{ tangent to the circle and radial acceleration } a_R \text{ pointing toward the center.}$ $\text{Step 2: Tangential acceleration is zero.}$ $a_T = \frac{dv}{dt} = 0$ $\text{Step 3: Calculate the centripetal acceleration.}$ $a_R = \frac{v^2}{R} = \frac{2^2}{2} = 2 \text{ m/s}^2$

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
}