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Current Question (ID: 9581)

Question:
$\text{A body is moving with a velocity of 30 m/s towards the east. After 10 s, its velocity becomes 40 m/s towards the north. The average acceleration of the body is:}$
Options:
  • 1. $7 \text{ m/s}^2$
  • 2. $\sqrt{7} \text{ m/s}^2$
  • 3. $5 \text{ m/s}^2$
  • 4. $1 \text{ m/s}^2$
Solution:
$\text{Hint: The average acceleration depends on the net change in velocity.}$ $\text{Step 1: Find the change in velocity.}$ $\text{The angle between east and north is } 90°.$ $|\vec{v}_f - \vec{v}_i| = \sqrt{v_f^2 + v_i^2}$ $\text{Step 2: Find the average acceleration.}$ $a = \frac{|\vec{v}_f - \vec{v}_i|}{\Delta t}$ $= \frac{\sqrt{30^2 + 40^2}}{10} = \frac{\sqrt{900 + 1600}}{10}$ $= 5 \text{ ms}^{-2}$ $\text{Hence, option (3) is the correct answer.}$

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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
}