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Current Question (ID: 18908)

Question:

$\text{The period of oscillation on a simple pendulum is } T = 2\pi \sqrt{\frac{L}{g}}.$ $\text{The measured value of } L \text{ is } 20.0 \text{ cm known to have } 1 \text{ mm accuracy and the time for } 100 \text{ oscillations of the pendulum is found to be } 90 \text{ s using a wristwatch of } 1 \text{ s resolution. The accuracy in the determination of } g \text{ is:}$

Options:

1. $2\%$
2. $3\%$
3. $1\%$
4. $5\%$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{L}{g}}$ $\text{Measured length } L = 20.0 \text{ cm}$ $\text{error in length } \Delta L = 1 \text{ mm} = 0.1 \text{ cm}$ $\text{For 100 oscillation time measured } t = 90 \text{ s}$ $\text{error in this measured time } \Delta t = 1 \text{ s}$ $t = nT \quad \text{...}(i)$ $(n = \text{no. of oscillation, } T=\text{time period})$ $\Delta t = n\Delta T \quad \text{...}(ii)$ $\text{Equation (ii)/(i)}$ $\frac{\Delta t}{t} = \frac{\Delta T}{T} \Rightarrow \frac{\Delta T}{T} = \frac{\Delta t}{t}$ $\Rightarrow \left( \frac{\Delta T}{T} \right) = \frac{1}{90}$ $\text{Now } T = 2\pi \sqrt{\frac{L}{g}}$ $g = \frac{4\pi^2 L}{T^2}$ $\frac{\Delta g}{g} = \left[ \frac{\Delta L}{L} + \frac{2\Delta T}{T} \right]$ $\frac{\Delta g}{g} = \left[ \frac{0.1}{20.0} + 2 \left( \frac{1}{90} \right) \right]$ $\% \text{ error } = \frac{\Delta g}{g} \times 100$ $\Rightarrow \left[ \frac{1}{200} + \frac{1}{45} \right] \times 100 = 2.72\% \approx 3\%$

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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
}