Import Question JSON

Current Question (ID: 9598)

Question:

$\text{A particle starts from the origin at } t = 0 \text{ with a velocity of } 5.0\hat{i} \text{ m/s and moves in the } x-y \text{ plane under the action of a force that produces a constant acceleration of } (3.0\hat{i} + 2.0\hat{j}) \text{ m/s}^2\text{. What is the speed of the particle at the instant its } x\text{-coordinate is 84 m?}$

Options:

1. $36 \text{ m/s}$
2. $26 \text{ m/s}$
3. $1 \text{ m/s}$
4. $0 \text{ m/s}$

Solution:

\text{Hint: Use the equations of motion.} \text{Step 1: Find the position of the particle after time t.} \text{As } \vec{r}(t) = 0 \text{ at } t = 0, \text{The position of the particle at time t is given by:} \vec{r}(t) = \vec{v}_0 t + \frac{1}{2}\vec{a}t^2 = 5.0\hat{i}t + \frac{1}{2}(3.0\hat{i} + 2.0\hat{j})t^2 = (5.0t + 1.5t^2)\hat{i} + 1.0t^2\hat{j} \text{Therefore,} x(t) = 5.0t + 1.5t^2 y(t) = 1.0t^2 \text{Step 2: Find t at which } x(t) = 84 \text{ m} \text{Given } x(t) = 84 \text{ m, } t = ? 5.0t + 1.5t^2 = 84 \Rightarrow t = 6 \text{ s} \text{Step 3: Find the velocity at } t = 6 \text{ s} \text{As } \vec{v} = \frac{d\vec{r}}{dt} = (5.0 + 3.0t)\hat{i} + 2.0t\hat{j} \text{So, at } t = 6 \text{ s,} \vec{v} = 23.0\hat{i} + 12.0\hat{j} \text{Speed} = |\vec{v}| = \sqrt{(23)^2 + (12)^2} \approx 26 \text{ ms}^{-1} \text{Hence, option (2) 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
}