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Current Question (ID: 8761)

Question:

$\text{A physical parameter 'a' can be determined by measuring the parameters } b, c, d \text{ and } e \text{ using the relation, } a = \frac{b^\alpha c^\beta}{d^\gamma e^\delta}. \text{ If the maximum errors in the measurement of } b, c, d, \text{ and } e \text{ are } b_1\%, c_1\%, d_1\% \text{ and } e_1\% \text{, then the maximum error in the value of 'a' determined by the experiment is:}$

Options:

1. $(b_1 + c_1 + d_1 + e_1)\%$
2. $(b_1 + c_1 - d_1 - e_1)\%$
3. $(\alpha b_1 + \beta c_1 - \gamma d_1 - \delta e_1)\%$
4. $(\alpha b_1 + \beta c_1 + \gamma d_1 + \delta e_1)\%$

Solution:

$\text{Hint: Error is always additive in nature.}$ $\text{Step: Find the maximum error in } a. a = \frac{b^\alpha c^\beta}{d^\gamma e^\delta}$ $\text{So the maximum error in 'a' is given by;}$ $\left( \frac{\Delta a}{a} \times 100 \right)_{\text{max}} = \alpha \cdot \left( \frac{\Delta b}{b} \times 100 \right) + \beta \cdot \left( \frac{\Delta c}{c} \times 100 \right) + \gamma \cdot \left( \frac{\Delta d}{d} \times 100 \right) + \delta \cdot \left( \frac{\Delta e}{e} \times 100 \right)$ $\left( \frac{\Delta a}{a} \times 100 \right)_{\text{max}} = (\alpha b_1 + \beta c_1 + \gamma d_1 + \delta e_1)\%$ $\text{Hence, option (4) is the correct answer.}$

Import JSON File

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
}