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Current Question (ID: 12059)

Question:

A body travels uniformly a distance of $(13.8\pm0.2)\text{ m}$ in a time $(4.0\pm0.3)\text{ s}$. The velocity of the body within error limits is

Options:

1. $(3.45\pm0.2)ms^{-1}$
2. $(3.45\pm0.3)ms^{-1}$
3. $(3.45\pm0.4)ms^{-1}$
4. $(3.45\pm0.5)ms^{-1}$

Solution:

First, calculate the velocity: $v = \frac{s}{t} = \frac{13.8}{4.0} = 3.45\text{ m/s}$. To find the error in velocity, we use the percentage error formula for division: $\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t}$. $\%\text{ error in s} = \frac{0.2}{13.8}\times100 \approx 1.45\%$. $\%\text{ error in t} = \frac{0.3}{4.0}\times100 = 7.5\%$. Total percentage error in velocity = $1.45\% + 7.5\% = 8.95\%$. This is incorrect based on the options. Let's re-evaluate using the absolute error. $\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t} = \frac{0.2}{13.8} + \frac{0.3}{4.0} \approx 0.0145 + 0.075 = 0.0895$. $\Delta v = v \times 0.0895 = 3.45 \times 0.0895 \approx 0.3087\text{ m/s}$. Rounding this value gives us $(3.45 \pm 0.3)\text{ m/s}$.

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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
}