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Current Question (ID: 8762)

Question:
$\text{A physical quantity } P \text{ is given by } P = \frac{A^3 B^{1/2}}{C^{-4} D^{3/2}}. \text{ The quantity which contributes the maximum percentage error in } P \text{ is:}$
Options:
  • 1. $A$
  • 2. $B$
  • 3. $C$
  • 4. $D$
Solution:
$\text{Hint: Quantity with maximum power contributes to maximum error.}$ $\text{Explanation: The physical quantity } P \text{ is given by; } P = \frac{A^3 B^{1/2}}{C^{-4} D^{3/2}}.$ $\text{In this equation, quantity } C \text{ has the highest power, which means it contributes the most significant error to } P.$ $\text{Hence, option (3) is the correct answer.}$

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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
}