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Current Question (ID: 9602)

Question:

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

Options:

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

Solution:

$\text{Hint: Use equations of motion.}$ $\text{Step 1: Find the position of the particle after time t.}$ $\text{Here as } r(t = 0) = 0$ $\text{So, the position of the particle at time t is given by:}$ $r(t) = v_0 t + \frac{1}{2}at^2$ $= 5.0\hat{i}t + (1/2)(3.0\hat{i} + 2.0\hat{j})t^2$ $= (5.0t + 1.5t^2)\hat{i} + 1.0t^2\hat{j}$ $\text{Therefore,}$ $x(t) = 5.0t + 1.5t^2$ $y(t) = 1.0t^2$ $\text{Step 2: Find time at which } x(t) = 84 \text{ m.}$ $\text{Given } x(t) = 84 \text{ m, } t = ?$ $5.0t + 1.5t^2 = 84$ $\Rightarrow t = 6 \text{ sec}$ $\text{Step 3: Find } y(t = 6 \text{ sec}).$ $\text{At } t = 6 \text{ s, } y = 1.0 \times (6)^2 = 36.0 \text{ m}$

Import JSON File

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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
}