Import Question JSON

Current Question (ID: 19896)

Question:

$\text{A body of mass } M \text{ and a charge } q \text{ is connected to a spring of spring constant } k. \text{ It oscillates along the } x\text{-direction about its equilibrium position, taken to be at } x = 0, \text{ with an amplitude } A. \text{ An electric field } E \text{ is applied along the } x\text{-direction.}$ $\text{Which of the following statements is correct?}$

Options:

1. $\text{The total energy of the system is } \frac{1}{2} m \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}.$
2. $\text{The new equilibrium position is at a distance } \frac{2qE}{k} \text{ from } x = 0.$
3. $\text{The new equilibrium position is at a distance } \frac{qE}{2k} \text{ from } x = 0.$
4. $\text{The total energy of the system is } \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} \frac{q^2 E^2}{k}.$

Solution:

$\text{Hint: } X_{\text{equi}} = \frac{qE}{k}$ $\text{Step 1: Draw the diagram.}$ $x = 0$ $M, q$ $E$ $qE$ $k x_0$ $M, q$ $\text{Step 2: Find the total energy of the system at the mean position.}$ $T. E. = K_{\text{mean position}} + U_{\text{mean position}}$ $\Rightarrow T. E. = \frac{1}{2} m v_{\text{max}}^2 + \frac{1}{2} k x_0^2$ $\Rightarrow T. E. = \frac{1}{2} m (A \omega)^2 + \frac{1}{2} k \left( \frac{qE}{k} \right)^2 \quad [\because kx_0 = qE]$ $\Rightarrow T. E. = \frac{1}{2} m \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}