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Current Question (ID: 16436)

Question:

$\text{In the Millikan oil drop experiment, a charged drop falls with a terminal velocity } v. \text{ If an electric field } E \text{ is applied vertically upwards it moves with terminal velocity } 2v \text{ in the upward direction. If the electric field reduces to } \frac{E}{2} \text{ then its terminal velocity will be:}$

Options:

1. $\frac{v}{2}$
2. $v$
3. $\frac{3v}{2}$
4. $2v$

Solution:

$\text{Hint: } F_v = mg = 6 \pi n r v_0$ $\text{Step: Find the terminal velocity.}$ $F_v = mg = 6 \pi n r v_0 \quad \cdots (1)$ $\text{when the field is applied vertically upward}$ $qE = mg + 6 \pi n r (2v_0) \quad \cdots (2)$ $\text{From (1) and (2)}$ $qE = 18 \pi n r v_0 \quad \cdots (3)$ $\text{If the field is reduced to half } mg + 6 \pi n r v' = \frac{qE}{2} \text{ from (3) putting the value of } qE$ $6 \pi n r v_0 + 6 \pi n r v' = \frac{18 \pi n r v_0}{2}$ $6 \pi n r v' = 3 \pi n r v_0$ $\Rightarrow v' = \frac{v_0}{2}$ $\text{Hence, option (1) is the correct answer.}$

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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
}