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Current Question (ID: 18948)

Question:
$\text{What is the maximum percentage error in the measurement of the quantity}$ $I = \frac{a^2b^3}{c\sqrt{d}},$ $\text{if the percentage errors in the measurements of } a, b, c, \text{ and } d \text{ are } 1\%, 2\%, 3\%, \text{ and } 4\%, \text{ respectively?}$
Options:
  • 1. 11\%
  • 2. 12\%
  • 3. 9\%
  • 4. 13\%
Solution:
$\text{Hint: Use the concept of relative error.}$ $\text{Step: Find the maximum percentage error in the measurement of } I.$ $\text{The quantity } I \text{ is defined as;}$ $I = \frac{a^2b^3}{c\sqrt{d}}$ $\text{The maximum percentage error in the measurement of } I \text{ is given by;}$ $\frac{\Delta I}{I} \times 100 = 2 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + \left( \frac{\Delta c}{c} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta d}{d} \times 1 \right)$ $\Rightarrow \frac{\Delta I}{I} \times 100 = 2 \times 1\% + 3 \times 2\% + 3\% + \frac{1}{2} \times 4\%$ $\Rightarrow \frac{\Delta I}{I} \times 100 = 2\% + 6\% + 3\% + 2\%$ $\Rightarrow \frac{\Delta I}{I} \times 100 = 13\%$ $\text{Hence, option (4) is the correct answer.}$

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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
}