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Current Question (ID: 19391)

Question:

$\text{A force } \vec{F} = -c\hat{k} \text{ is acting on a particle at the origin. The torque of this force about the point } (-2, 3, 0) \text{ is given by } -c(A\hat{i} + B\hat{j}).$ $\text{If } \frac{A}{B} = \frac{x}{2}, \text{ then the value of } x \text{ is:}$

Options:

1. $\frac{2}{3}$
2. $3$
3. $1.5$
4. $1$

Solution:

$\text{Hint: } \vec{\tau} = \vec{r} \times \vec{F}$ $\text{Step: Find the value of } x.$ $\text{Given: The torque of the given force about point } (-2, 3, 0) \text{ is } -c(A\hat{i} + B\hat{j}).$ $\vec{\tau} = \vec{r} \times \vec{F}$ $\Rightarrow -c(A\hat{i} + B\hat{j}) = \left((0 - (-2))\hat{i} + (0 - 3)\hat{j}\right) \times (-c\hat{k})$ $\Rightarrow (A\hat{i} + B\hat{j}) = (2\hat{i} - 3\hat{j}) \times \hat{k}$ $\Rightarrow (A\hat{i} + B\hat{j}) = -2\hat{j} - 3\hat{i} \left[\hat{i} \times \hat{k} = -\hat{j}, \hat{j} \times \hat{k} = \hat{i}\right]$ $\text{Comparing both sides we get;} A = -3, \ B = -2$ $\text{The required ratio is } \frac{A}{B} = \frac{-3}{-2} = \frac{3}{2} = \frac{x}{2}$ $\text{Therefore, the value of } x \text{ is } 3.$ $\text{Hence, option (2) is the correct answer.}$

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
}