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Current Question (ID: 20025)

Question:

$\text{A charged particle (mass } m \text{ and charge } q) \text{ moves along X axis with velocity } V_0. \text{ When it passes through the origin it enters a region having uniform electric field } \vec{E} = -E\hat{j} \text{ which extends upto } x = d. \text{ Equation of path of electron in the region } x > d \text{ is:}$

Options:

1. $y = \frac{qEd}{mV_0^2} \left( \frac{d}{2} - x \right)$
2. $y = \frac{qEd}{mV_0^2} (x - d)$
3. $y = \frac{qEd}{mV_0^2} x$
4. $y = \frac{qEd^2}{mV_0^2} x$

Solution:

$\text{Hint: Horizontal component of velocity remains constant.}$ $\text{Let particle have charge } q \text{ and mass } m$ $\text{Solve for } (q, m) \text{ mathematically}$ $F_x = 0, a_x = 0, (v)_x = \text{constant}$ $\text{Time taken to reach at } 'P' = \frac{d}{v_0} = t_0$ $\left( \text{Along } -y \right), y_0 = 0 + \frac{1}{2} \cdot \frac{qE}{m} \cdot t_0^2$ $v_x = v_0$ $v = u + at$ $\text{speed } v_{y_0} = \frac{qE}{m} \cdot t_0$ $\tan \theta = \frac{v_y}{v_x} = \frac{qEt_0}{m \cdot v_0}, \left( t_0 = \frac{d}{v_0} \right)$ $\tan \theta = \frac{qEd}{m \cdot v_0^2}$ $\text{Slope} = \frac{-qEd}{m \cdot v_0^2}$ $\text{Now we have to find eqn of straight line whose slope is } \frac{-qEd}{m \cdot v_0^2} \text{ and it pass through point } \rightarrow (d, -y_0) \text{ Because after } x > d$ $\text{No electric field } = F_{net} = 0, \vec{v} = m x + c, \{ m = \frac{qEd}{m \cdot v_0^2} \}$ $-\frac{qEd}{m \cdot v_0^2} \cdot d + c = c = y_0 + \frac{qEd^2}{m \cdot v_0^2} (d, -$ $\text{Put the value}$ $y = \frac{-qEd}{m \cdot v_0^2} x - y_0 + \frac{qEd^2}{m \cdot v_0^2}$ $y_0 = \frac{1}{2} \cdot \frac{qE}{m} \left( \frac{d}{v_0} \right)^2 = \frac{qEd^2}{2m \cdot v_0^2}$ $y_0 = \frac{qEd^2}{2m \cdot v_0^2}$ $y = \frac{-qEd}{m \cdot v_0^2} x + \frac{qEd^2}{2m \cdot v_0^2}$ $y = \frac{qEd}{m \cdot v_0^2} \left( \frac{d}{2} - x \right)$

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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
}