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Current Question (ID: 19078)

Question:

$\text{The position vector of a particle changes with time according to the relation,}$ $\vec{r}(t) = (15t^2)\hat{i} + (4 - 20t^2)\hat{j}, \text{ where } \vec{r}(t) \text{ is in metres and } t \text{ is in seconds.}$ $\text{What is the magnitude of the acceleration at } t = 1 \text{ second?}$

Options:

1. $100 \text{ m/s}^2$
2. $40 \text{ m/s}^2$
3. $50 \text{ m/s}^2$
4. $25 \text{ m/s}^2$

Solution:

$\text{Hint: } a = \frac{d^2\vec{r}}{dt^2}$ $\text{Step: Find the acceleration of the particle at } t = 1 \text{ s.}$ $\text{The position vector of the particle is given as;}$ $\vec{r}(t) = (15t^2)\hat{i} + (4 - 20t^2)\hat{j},$ $\text{The velocity of the particle is given by;}$ $\vec{v} = \frac{d\vec{r}}{dt} = 30t\hat{i} - 40t\hat{j}$ $\text{The acceleration of the particle is given by;}$ $\vec{a} = \frac{d\vec{v}}{dt} = 30\hat{i} - 40\hat{j}$ $\text{The magnitude of the acceleration of the particle is given by;}$ $a = \sqrt{(30)^2 + (40)^2} = \sqrt{2500} = 50 \text{ m/s}^2$ $\text{Hence, option (3) is the correct answer.}$

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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
}