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Current Question (ID: 16483)

Question:

$\text{Figure shows a ball having a charge } q \text{ fixed at a point } A. \text{ Two identical balls having charges } +q \text{ and } -q \text{ and mass } 'm' \text{ each are attached to the ends of a light rod of length } 2a. \text{ The rod is free to rotate about a fixed axis perpendicular to the plane of the paper and passing through the mid-point of the rod. The system is released from the situation as shown in the figure. The angular velocity of the rod when the rod becomes horizontal will be:}$

Options:

1. $\frac{\sqrt{2}q}{3\pi \varepsilon_0 ma^3}$
2. $\frac{q}{\sqrt{3\pi \varepsilon_0 ma^3}}$
3. $\frac{q}{\sqrt{6\pi \varepsilon_0 ma^3}}$
4. $\frac{\sqrt{2}q}{4\pi \varepsilon_0 ma^3}$

Solution:

$\text{(3) There will be a net torque on the rod which rotates the rod in the anti-clockwise direction. The loss in electrostatic potential energy will be equal to the gain in rotational kinetic energy.}$ $\text{Initial potential energy of the system } = 0 \quad \text{(i)}$ $\text{The rod will rotate in anti-clockwise sense.}$ $\text{When rod becomes horizontal}$ $U_f = -\frac{Kq^2}{a} + \frac{Kq^2}{3a} = -\frac{2Kq^2}{3a} \quad \text{(ii)}$ $\text{loss in P.E. } = \frac{2Kq^2}{3a} = \text{gain in K.E.}$ $\therefore \frac{2Kq^2}{3a} = 2 \times \frac{1}{2} \times ma^2 \omega^2 \quad \text{(using) } \frac{1}{2} I \omega^2$ $\omega^2 = \frac{q^2}{6\pi \varepsilon_0 ma^3}$ $\therefore \omega = \frac{q}{\sqrt{6\pi \varepsilon_0 ma^3}}$

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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
}