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Current Question (ID: 9557)

Question:

$\text{A person moved from A to B on a circular path as shown in the figure below. If the distance travelled by him is 60 m, then the magnitude of displacement would be:}$ $\text{(given } \cos 135° = -0.7\text{)}$

Options:

1. $42 \text{ m}$
2. $47 \text{ m}$
3. $19 \text{ m}$
4. $40 \text{ m}$

Solution:

$\text{Hint: } R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ $\text{Step 1: Find the radius of the circle path.}$ $\Rightarrow \frac{\theta}{360°} \times 2\pi R = 60$ $\Rightarrow \frac{3 \times 45}{360} \times 2\pi R = 60$ $\Rightarrow \frac{\pi R}{8} = 10$ $\Rightarrow R = \frac{80}{\pi}$ $\text{Step 2: Find the magnitude of the displacement of the person.}$ $\vec{R_1} + \overrightarrow{AB} = \vec{R_2}$ $\Rightarrow \overrightarrow{AB} = \vec{R_2} - \vec{R_1}$ $|\overrightarrow{AB}| = |\vec{R_2} - \vec{R_1}|$ $\Rightarrow |\overrightarrow{AB}| = \sqrt{R^2 + R^2 - 2R^2 \times (-0.7)}$ $\Rightarrow |\overrightarrow{AB}| = \sqrt{2R^2 + 1.4R^2} = \sqrt{3.4R}$ $\Rightarrow |\overrightarrow{AB}| = \frac{80}{3.14} \times 1.82 = 47 \text{ m } [R = \frac{80}{\pi}]$ $\text{Hence, option (2) is the correct answer.}$

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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
}