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Current Question (ID: 20127)

Question:

$\text{A solid sphere of radius } R \text{ carries a charge } (Q + q) \text{ distributed uniformly over its volume.}$ $\text{A very small point like piece of it of mass } m \text{ gets detached from the bottom of the sphere and falls down vertically under gravity.}$ $\text{This piece carries charge } q. \text{ If it acquires a speed } v \text{ when it has fallen through a vertical height } y \text{ (see figure), then:}$ $\text{(assume the remaining portion to be spherical).}$

Options:

1. $v^2 = 2y \left[ \frac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g \right]$
2. $v^2 = y \left[ \frac{qQ}{4\pi\varepsilon_0 R^2 ym} + g \right]$
3. $v^2 = 2y \left[ \frac{qQ R}{4\pi\varepsilon_0 (R+y)^3 m} + g \right]$
4. $v^2 = y \left[ \frac{qR}{4\pi\varepsilon_0 R(R+y)m} + g \right]$

Solution:

$\text{Hint: Apply the law of conservation of energy.}$ $\frac{kQq}{R} + mgy = \frac{kQq}{R+y} + \frac{1}{2}mv^2$ $v^2 = 2gy + \frac{2kQqy}{mR(R+y)}$

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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
}