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Current Question (ID: 16608)

Question:

$\text{Drift velocity } v_d \text{ varies with the intensity of the electric field as per the relation:}$ $1. \ v_d \propto E$ $2. \ v_d \propto \frac{1}{E}$ $3. \ v_d = \text{constant}$ $4. \ v_d \propto E^2$

Options:

1. $v_d \propto E$
2. $v_d \propto \frac{1}{E}$
3. $v_d = \text{constant}$
4. $v_d \propto E^2$

Solution:

$\text{Hint: } v_d = \frac{e}{m} \times \frac{V}{lt}$ $\text{Step: Find the intensity of the electric field.}$ $\text{The drift velocity can be expressed using the formula:}$ $v_d = \frac{e}{m} \times \frac{V}{lt}$ $\text{The potential difference } V \text{ can be expressed in terms of the electric field } E$ $\text{and the length } L \text{ of the conductor:}$ $V = E \times L$ $\text{From the simplified equation, we can see that;}$ $\Rightarrow v_d \propto E$ $\text{Hence, option (1) is the correct answer.}$

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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
}