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Current Question (ID: 8718)

Question:

$\text{The dimensional formula of modulus of rigidity is:}$

Options:

1. $[ML^{-1}T^{-2}]$
2. $[ML^{-2}T^{2}]$
3. $[MLT^{-1}]$
4. $[MLT^{-2}]$

Solution:

\text{Hint: Modulus of rigidity is the ratio of shear stress to the shear strain.} \text{Explanation:} \text{Modulus of rigidity (shear modulus) is defined as the ratio of shear stress to shear strain.} \text{Shear stress has dimensions of pressure: } [ML^{-1}T^{-2}] \text{Shear strain is dimensionless (ratio of displacement to length).} \text{Therefore, modulus of rigidity} = \frac{\text{Shear stress}}{\text{Shear strain}} = \frac{[ML^{-1}T^{-2}]}{[1]} = [ML^{-1}T^{-2}]

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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
}