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Current Question (ID: 9573)

Question:

$\text{Three particles are moving with constant velocities } v_1, v_2 \text{ and } v \text{ respectively as given in the figure. After some time, if all the three particles are in the same line, then the relation among } v_1, v_2 \text{ and } v \text{ is:}$

Options:

1. $v = v_1 + v_2$
2. $v = \sqrt{v_1 v_2}$
3. $v = \frac{v_1 v_2}{v_1 + v_2}$
4. $v = \frac{\sqrt{2v_1 v_2}}{v_1 + v_2}$

Solution:

\text{Hint: Slope} = \frac{y_2 - y_1}{x_2 - x_1} \text{Step: Find the relation between the velocities of the particles.} \text{At time } t\text{, the positions of the particles are:} \text{A at } (0, v_2 t) \text{B at } \left(\frac{v_1 t}{\sqrt{2}}, \frac{v_1 t}{\sqrt{2}}\right) \text{C at } (v_1 t, 0) \text{At time } t\text{, A, B and C are on a straight line, so Slope of AC = Slope of BC} \frac{v_2 t - 0}{0 - v_1 t} = \frac{\frac{v_1 t}{\sqrt{2}} - 0}{\frac{v_1 t}{\sqrt{2}} - v_1 t} \frac{v_2}{-v_1} = \frac{\frac{v_1}{\sqrt{2}}}{\frac{v_1}{\sqrt{2}} - v_1} \frac{v_2}{v_1} = \frac{v_1}{\sqrt{2} \cdot v_1 - v_1} v_1 = \frac{\sqrt{2} v_1 v_2}{v_1 + v_2} \text{Therefore, the correct relation between the velocities of the particles is } v_1 = \frac{\sqrt{2} v_1 v_2}{v_1 + v_2} \text{Hence, option (4) 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
}