Import Question JSON

Current Question (ID: 16427)

Question:

$\text{A toy car with charge } q \text{ moves on a frictionless horizontal plane surface under the influence of a uniform electric field } \vec{E}. \text{ Due to the force } q\vec{E}, \text{ its velocity increases from } 0 \text{ to } 6 \text{ m/s in a one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between } 0 \text{ to } 3 \text{ seconds are respectively:}$

Options:

1. $2 \text{ m/s, } 4 \text{ m/s}$
2. $1 \text{ m/s, } 3 \text{ m/s}$
3. $1 \text{ m/s, } 3.5 \text{ m/s}$
4. $1.5 \text{ m/s, } 3 \text{ m/s}$

Solution:

$\text{The acceleration is given by;} \ a = \frac{6-0}{1} = 6 \text{ ms}^{-2}$ $\text{For } t = 0 \text{ to } t = 1 \text{ s,}$ $s = ut + (1/2)at^2$ $S_1 = \frac{1}{2} \times 6(1)^2 = 3 \text{ m} \quad \cdots (1)$ $\text{For } t = 1 \text{ s to } t = 2 \text{ s,}$ $S_2 = 6 \times 1 - \frac{1}{2} \times 6(1)^2 = 3 \text{ m}$ $\text{For } t = 2 \text{ s to } t = 3 \text{ s,}$ $S_3 = 0 - \frac{1}{2} \times 6(1)^2 = -3 \text{ m}$ $\text{The total displacement } S = S_1 + S_2 + S_3 = 3 \text{ m}$ $\text{The average velocity } v_{\text{avg}} = \frac{3}{3} = 1 \text{ ms}^{-1}$ $\text{The total distance travelled } = 9 \text{ m}$ $\text{The average speed } v_{\text{avg}} = \frac{9}{3} = 3 \text{ ms}^{-1}$ $\text{Hence, option (2) is the correct answer.}$

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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
}