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Current Question (ID: 20057)

Question:
$\text{An electric dipole is shown in the figure. If it is displaced by a small angle with respect to the electric field, the angular frequency of its oscillation is given by:}$
Options:
  • 1. $\sqrt{\frac{6qE}{ml}}$
  • 2. $\sqrt{\frac{3qE}{ml}}$
  • 3. $\sqrt{\frac{2qE}{ml}}$
  • 4. $\sqrt{\frac{qE}{ml}}$
Solution:
$\text{Hint: } \tau = qE \sin \theta$ $\text{Step 1: Find the moment of inertia about centre of mass i.e., } I_{CM}. \text{The moment of inertia of a given configuration about centre of mass is given by;}$ $I_{CM} = \frac{m}{2} r_1^2 + mr_2^2$ $\text{Where, } r_1 = \frac{l \times m}{m + m} = \frac{2l}{3}, r_2 = \frac{l \times \frac{m}{2}}{\frac{m}{2} + m} = \frac{l}{3}$ $\Rightarrow I_{CM} = \frac{m}{2} \left( \frac{2l}{3} \right)^2 + m \left( \frac{l}{3} \right)^2$ $\Rightarrow I_{CM} = \frac{ml^2}{3}$ $\text{Step 2: Find the angular frequency of oscillation.}$ $\text{We know that when a dipole is placed in a uniform electric field then torque acting on the dipole is given by;}$ $\vec{\tau} = \vec{p} \times \vec{E} = pE \sin \theta$ $\text{As } \theta \text{ is very small, } \sin \theta \approx \theta$ $\Rightarrow \vec{\tau} = qlE \theta$ $\Rightarrow I_{CM} \times \alpha = qlE \theta$ $\Rightarrow \alpha = \frac{qlE}{I_{CM}} \theta \quad \cdots (1)$ $\text{The general equation of angular SHM is given by;}$ $\alpha = \omega^2 \theta \quad \cdots (2)$ $\text{From equation (1) and (2) we get;}$ $\omega^2 = \frac{qlE}{I_{CM}}$ $\Rightarrow \omega = \sqrt{\frac{qlE}{\frac{ml^2}{3}}} = \sqrt{\frac{3qE}{ml}}$ $\text{Hence, option (2) is the correct answer.}$

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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
}