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Current Question (ID: 17300)

Question:
$L$, $C$ \text{ and } $R$ \text{ represent physical quantities inductance, capacitance and resistance respectively. The combination representing the dimension of frequency will be:}$ $1. \ LC$ $2. \ (LC)^{-\frac{1}{2}}$ $3. \ \left(\frac{L}{C}\right)^{-\frac{1}{2}}$ $4. \ \frac{C}{L}$
Options:
  • 1. $LC$
  • 2. $(LC)^{-\frac{1}{2}}$
  • 3. $\left(\frac{L}{C}\right)^{-\frac{1}{2}}$
  • 4. $\frac{C}{L}$
Solution:
$(2) \ \text{Frequency} = \frac{1}{2\pi\sqrt{LC}}$ $\text{So the combination which represents dimension of frequency is}$ $\frac{1}{\sqrt{LC}} = (LC)^{-1/2}$

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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
}