Import Question JSON

Current Question (ID: 9639)

Question:

$\text{A body is projected at an angle of 30° with the horizontal with a speed of 30 m/s. What is the angle made by the velocity vector with the horizontal after 1.5 sec? } (g = 10 \text{ m/s}^2)$

Options:

1. $0°$
2. $30°$
3. $60°$
4. $90°$

Solution:

$\text{Hint: Horizontal velocity remains constant.}$ $\text{Step 1: Find initial horizontal and vertical velocities.}$ $u_x = 30 \cos 30°$ $u_y = 30 \sin 30°$ $= 15 \text{ m/s}$ $\text{Step 2: Find vertical velocity after 1.5 sec.}$ $v_y = u + at$ $= 15 - 10 \times 1.5$ $= 0$ $\text{So, the velocity vector is parallel to the horizontal after 1.5 sec.}$ $\text{Hence, option (1) is the correct answer.}$

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
}