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Current Question (ID: 18973)

Question:
$\text{A physical quantity } z \text{ depends on four observables } a, b, c \text{ and } d, \text{ as}$ $z = \frac{a^2 b^{\frac{3}{2}}}{\sqrt{cd^3}}$ $\text{The percentage of error in the measurement of } a, b, c \text{ and } d \text{ is } 2\%, 1.5\%, 4\% \text{ and } 2.5\% \text{ respectively. The percentage of error in } z \text{ is:}$
Options:
  • 1. 12.25\%
  • 2. 14.5\%
  • 3. 16.5\%
  • 4. 13.5\%
Solution:
$\text{Hint: } \delta a = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100$

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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
}