Import Question JSON

Current Question (ID: 19091)

Question:

$\text{A person moves from point } A \text{ to point } B \text{ along a circular arc that}$ $\text{subtends an angle of } 135^\circ \text{ at the centre } O, \text{ as shown in the figure.}$ $\text{The length of the arc } AB \text{ is } 60 \text{ m. If } \cos 135^\circ = -0.7, \text{ the}$ $\text{magnitude of the displacement of the person is:}$

Options:

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

Solution:

$\text{Hint: } R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$ $\text{Step 1: Find the radius of the circular path.}$ $\text{The distance travelled by the body on the circular path is given by:}$ $\frac{\theta}{360^\circ} \times 2\pi R = 60$ $\Rightarrow \frac{135^\circ}{360^\circ} \times 2\pi R = 60$ $\Rightarrow R = \frac{80}{\pi}$ $\text{Step 2: Find the magnitude of the displacement of the person.}$ $\vec{R}_2 = \vec{R}_1 + \vec{AB}$ $\Rightarrow \vec{AB} = \vec{R}_2 - \vec{R}_1$ $\Rightarrow |\vec{AB}| = |\vec{R}_2 - \vec{R}_1|$ $\Rightarrow |\vec{AB}| = \sqrt{R^2 + R^2 - 2R^2 \times \cos 135^\circ}$ $\Rightarrow |\vec{AB}| = \sqrt{R^2 + R^2 - 2R^2 \times (-0.7)}$ $\Rightarrow |\vec{AB}| = \sqrt{2R^2 + 1.4R^2}$ $\Rightarrow |\vec{AB}| = \sqrt{3.4R^2}$ $\Rightarrow |\vec{AB}| = 1.84 \times \frac{80}{3.14}$ $\Rightarrow |\vec{AB}| = 46.9 \approx 47 \text{ m}$ $\text{Hence, option (2) is the correct answer.}$

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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
}