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Current Question (ID: 8766)

Question:

\text{In an experiment, the following observations were recorded: initial length } L = 2.820 \text{ m, mass } M = 3.00 \text{ kg, change in length } \Delta l = 0.087 \text{ cm, diameter } D = 0.041 \text{ cm. Taking } g = 9.81 \text{ m/s}^2 \text{ and using the formula } Y = \frac{4MgL}{\pi D^2 \Delta l}\text{, the maximum permissible error in } Y \text{ will be:}

Options:

1. $7.96\%$
2. $4.56\%$
3. $6.50\%$
4. $8.42\%$

Solution:

\text{Hint: } \frac{\Delta Y}{Y} \times 100 = \left( \frac{\Delta M}{M} + \frac{\Delta g}{g} + \frac{\Delta L}{L} + \frac{2\Delta D}{D} + \frac{\Delta l}{l} \right) \times 100 \text{Step: Find the maximum possible error in } Y\text{.} \text{The expression of Young's modulus is given by: } Y = \frac{4MgL}{\pi D^2 l} \text{So, the maximum permissible error in } Y \text{ is given by:} \frac{\Delta Y}{Y} \times 100 = \left( \frac{\Delta M}{M} + \frac{\Delta g}{g} + \frac{\Delta L}{L} + \frac{2\Delta D}{D} + \frac{\Delta l}{l} \right) \times 100 \frac{\Delta Y}{Y} \times 100 = \left( \frac{1}{300} + \frac{1}{981} + \frac{1}{2820} + 2 \times \frac{1}{41} + \frac{1}{87} \right) \times 100 \frac{\Delta Y}{Y} \times 100 = 0.065 \times 100 = 6.5\% \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
}