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Current Question (ID: 8763)

Question:

$\text{The length of a cylinder is measured with a meter rod having the least count of 0.1 cm. Its diameter is measured with vernier calipers having the least count of 0.01 cm. Given that the length is 5.0 cm and the radius is 2.0 cm. The percentage error in the calculated value of the volume will be:}$

Options:

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

Solution:

$\text{Hint: Use error analysis.}$ $\text{Step: Find the maximum error in volume.}$ $\text{The volume of cylinder is given by; } V = \pi r^2 L$ $\text{Percentage error in volume:}$ $\frac{\Delta V}{V} \times 100 = \frac{2\Delta r}{r} \times 100 + \frac{\Delta L}{L} \times 100$ $\Rightarrow \frac{\Delta V}{V} \times 100 = \left( 2 \times \frac{0.01}{2.0} \times 100 + \frac{0.1}{5.0} \times 100 \right) = (1 + 2)\% = 3\%$ $\text{Hence, option (3) is the correct answer.}$

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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
}