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Current Question (ID: 20022)

Question:

$\text{A simple pendulum of length } L \text{ is suspended between the plates of a}$ $\text{parallel-plate capacitor producing a uniform electric field } E, \text{ as}$ $\text{shown in the figure. The bob has a mass } m \text{ and a charge } q. \text{ The time}$ $\text{period of oscillation of the pendulum is given by:}$

Options:

1. $2\pi \sqrt{\frac{L}{\sqrt{g^2 - \frac{q^2E^2}{m^2}}}}$
2. $2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$
3. $2\pi \sqrt{\frac{L}{g - \frac{qE}{m}}}$
4. $2\pi \sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}\right)^2}}}$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}}$ $\text{Step 1: Draw the diagram.}$ $\text{Step 2: Find the net acceleration of the bob.}$ $F_{\text{net}} = \sqrt{(mg)^2 + (qE)^2}$ $\Rightarrow ma_{\text{net}} = \sqrt{(mg)^2 + (qE)^2}$ $\Rightarrow a_{\text{net}} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$ $\text{Step 3: Find the time period of the pendulum.}$ $T = 2\pi \sqrt{\frac{L}{a_{\text{net}}}}$ $\Rightarrow T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \left(\frac{qE}{m}\right)^2}}}$ $\text{Hence, option (4) is the correct answer.}$

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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
}