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Current Question (ID: 18922)

Question:

$\text{The density of a solid metal sphere is determined by measuring its mass and its diameter.}$ $\text{The maximum error in the density of the sphere is } \left( \frac{x}{100} \right)\%.$ $\text{If the relative errors in measuring the mass and the diameter are } 6.0\% \text{ and } 1.5\% \text{ respectively, the value of } x \text{ is:}$

Options:

1. 503
2. 1050
3. 532
4. 120

Solution:

$\text{Hint: } \rho = \frac{M}{V}$ $\text{Ans is } 1050.00$ $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi \left( \frac{D}{2} \right)^3}$ $\rho = \frac{6}{\pi} MD^{-3}$ $\text{Taking log}$ $\ln p = \ln \left( \frac{6}{\pi} \right) + \ln M - 3 \ln D$ $\text{Differentiates}$ $\frac{dp}{p} = 0 + \frac{dM}{M} - 3 \frac{d(D)}{D}$ $\text{For maximum error}$ $100 \times \frac{dp}{p} = \frac{dM}{M} \times 100 + 3 \frac{dD}{D} \times 100$ $= 6 + 3 \times 1.5$ $= 10.5\%$ $= \frac{1050}{100}\% \text{ so } x = 1050.00$

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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
}