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Current Question (ID: 8765)

Question:

$\text{The period of oscillation of a simple pendulum is given by } T = 2\pi\sqrt{\frac{L}{g}} \text{ where } L \text{ is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillations is measured by a stopwatch of least count 0.1 s. The percentage error in } g \text{ is:}$

Options:

1. $0.1\%$
2. $1\%$
3. $0.2\%$
4. $0.8\%$

Solution:

$\text{Hint: } T = 2\pi\sqrt{\frac{L}{g}}$ $\text{Step: Find the error in acceleration due to gravity.}$ $\text{The time period of the simple pendulum is given by;}$ $T = 2\pi\sqrt{\frac{L}{g}}$ $\Rightarrow T^2 = 4\pi^2 \frac{L}{g}$ $\Rightarrow g = \frac{4\pi^2 L}{T^2}$ $\% \text{ error in } L = \frac{1 \text{ mm}}{100 \text{ cm}} \times 100 = \frac{0.1}{100} \times 100 = 0.1\%$ $\% \text{ error in } T = \frac{0.1}{2 \times 100} \times 100 = 0.05\%$ $\Rightarrow \% \text{ error in } g = \% \text{ error in } L + 2(\% \text{ error in } T) = 0.1 + 2 \times 0.05 = 0.2\%$ $\text{Hence, option (3) is the correct answer.}$

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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
}