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Current Question (ID: 18924)

Question:

$\text{The period of oscillation of a simple pendulum } T = 2\pi \sqrt{\frac{L}{g}}.$ $\text{Measured value of } 'L' \text{ is } 1.0 \text{ m from meter scale having a minimum division of } 1 \text{ mm and time of one complete oscillation is } 1.95 \text{ s measured from stopwatch of } 0.01 \text{ s resolution. The percentage error in the determination of } 'g' \text{ will be:}$

Options:

1. $1.13\%$
2. $1.03\%$
3. $1.33\%$
4. $1.30\%$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{l}{g}}$ $T = 2\pi \sqrt{\frac{\ell}{g}}$ $g = \frac{4\pi^2 \ell}{T^2}$ $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + \frac{2 \Delta T}{T}$ $\frac{\Delta g}{g} = \frac{1 \times 10^{-3}}{1} + 2 \times \frac{0.01}{1.95}$ $\frac{\Delta g}{g} = 0.0113 \text{ or } 1.13\%$ $\text{option } (1) \text{ is correct}$

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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
}