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Current Question (ID: 18986)

Question:

$\text{If momentum } [P], \text{ area } [A] \text{ and time } [T] \text{ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:}$

Options:

1. $[PA^{-1}T^0]$
2. $[PAT^{-1}]$
3. $[PA^{-1}T]$
4. $[PA^{-1}T^{-1}]$

Solution:

$\text{The dimensional formula for viscosity is } [ML^{-1}T^{-1}].$ $\text{Given } [P] = [MLT^{-1}], [A] = [L^2], [T] = [T].$ $\text{Substituting these into the formula for viscosity:}$ $[ML^{-1}T^{-1}] = [P][A]^{-1}[T]^0 = [PA^{-1}T^0].$

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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
}