Import Question JSON

Current Question (ID: 8768)

Question:

$\text{The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error would be, respectively}$

Options:

1. $164 \pm 3 \text{ cm}^2$
2. $163.62 \pm 2.6 \text{ cm}^2$
3. $163.6 \pm 2.6 \text{ cm}^2$
4. $163.62 \pm 3 \text{ cm}^2$

Solution:

$\text{Hint: The result should have as many significant figures as there are in the value with the least significant figures.}$ $\text{Step 1: Find the area of the sheet.}$ $\text{Given:}$ $\text{length } l = (16.2 \pm 0.1) \text{ cm}$ $\text{Breadth } b = (10.1 \pm 0.1) \text{ cm}$ $\text{Area } A = l \times b$ $= (16.2 \text{ cm}) \times (10.1 \text{ cm}) = 163.62 \text{ cm}^2$ $\text{Both measurements have 3 significant figures, so rounding off to three significant digits, area } A = 164 \text{ cm}^2$ $\text{Step 2: Find the error.}$ $\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1}$ $= \frac{1.01 + 1.62}{16.2 \times 10.1} = \frac{2.63}{163.62}$ $\Rightarrow \Delta A = A \times \frac{2.63}{163.62} = 163.62 \times \frac{2.63}{163.62} = 2.63 \text{ cm}^2$ $\Rightarrow \Delta A = 3 \text{ cm}^2 \text{ (By rounding off to one significant figure)}$ $\text{Therefore, area } A = A \pm \Delta A = (164 \pm 3) \text{ cm}^2$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

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
}