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Current Question (ID: 16945)

Question:
$\text{What is the dimensional formula of magnetic flux?}$
Options:
  • 1. $[ML^2T^{-2}A^{-1}]$
  • 2. $[MLT^{-1}A^{-2}]$
  • 3. $[ML^2T^{-3}A^{-1}]$
  • 4. $[ML^{-2}T^{-2}A^{-2}]$
Solution:
$\text{Hint: } \phi = BA \cos \theta$ $\text{Step: Find the dimensional formula of magnetic flux.}$ $\text{The magnetic flux is given by:}$ $\phi = BA \cos \theta$ $\text{The magnetic force is given by:}$ $\vec{F} = q(\vec{v} \times \vec{B})$ $\text{So, the dimensional formula of magnetic flux (}\phi\text{) is given by:}$ $\phi = \frac{[F][A]}{[q][v]}$ $\phi = \frac{[MLT^{-2}][L^2][AT^{-1}]}{[LT^{-1}]}$ $\Rightarrow \phi = [ML^2T^{-2}A^{-1}]$ $\text{Hence, option (1) is the correct answer.}$

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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
}