Import Question JSON

Current Question (ID: 19083)

Question:

$\text{A particle starts from the origin at time } t = 0 \text{ with an initial velocity of } 5\hat{j} \text{ ms}^{-1}. \text{ It moves in the } XY\text{-plane under a constant acceleration of } (10\hat{i} + 4\hat{j}) \text{ ms}^{-2}. \text{ At some later time } t, \text{ the coordinates of the particle are } (20 \text{ m}, y_0 \text{ m}). \text{ The values of } t \text{ and } y_0 \text{ are, respectively:}$ $1. \ 4 \text{ s and } 52 \text{ m}$ $2. \ 5 \text{ s and } 25 \text{ m}$ $3. \ 2 \text{ s and } 18 \text{ m}$ $4. \ 2 \text{ s and } 24 \text{ m}$

Options:

1. $4 \text{ s and } 52 \text{ m}$
2. $5 \text{ s and } 25 \text{ m}$
3. $2 \text{ s and } 18 \text{ m}$
4. $2 \text{ s and } 24 \text{ m}$

Solution:

$\text{Hint: } \vec{s} = \vec{u}t + \frac{1}{2} \vec{a}t^2$ $\text{Step 1: Find the value of time } t.$ $\text{Apply second equation of motion along horizontal direction:}$ $s_x = u_x t + \frac{1}{2} a_x t^2$ $\Rightarrow 20 = 0 + \frac{1}{2} \times 10t^2$ $\Rightarrow t = 2 \text{ s}$ $\text{Step 2: Find the value of } y_0.$ $\text{Apply second equation of motion along vertical direction:}$ $s_y = u_y t + \frac{1}{2} a_y t^2$ $\Rightarrow y_0 = 5 \times 2 + \frac{1}{2} \times 4 \times 2^2$ $\Rightarrow y_0 = 18 \text{ m}$ $\text{Hence, option } (3) \text{ is the correct answer.}$

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
}