Import Question JSON

Current Question (ID: 18934)

Question:

$\text{A student in the laboratory measured thickness of a wire using a screw gauge. The readings are } 1.22 \text{ mm, } 1.23 \text{ mm, } 1.19 \text{ mm and } 1.20 \text{ mm. The percentage error is } \frac{x}{121} \%. \text{ The value of } x \text{ is:}$

Options:

1. $100$
2. $200$
3. $150$
4. $300$

Solution:

$\text{Percentage error} = \frac{\text{Mean deviation}}{\text{Mean reading}} \times 100\%$ $\text{Step 1: Find the mean reading.}$ $\text{Mean Reading} = \frac{1.22 + 1.23 + 1.19 + 1.20}{4}$ $\Rightarrow \text{Mean Reading} = 1.21 \text{ mm}$ $\text{Step 2: Find the deviations from mean.}$ $\Delta t_1 = |1.22 - 1.21| = 0.01$ $\Delta t_2 = |1.23 - 1.21| = 0.02$ $\Delta t_3 = |1.19 - 1.21| = 0.02$ $\Delta t_4 = |1.20 - 1.21| = 0.01$ $\text{Step 3: Find the mean deviation.}$ $\Delta t = \frac{0.01 + 0.02 + 0.02 + 0.01}{4}$ $\Delta t = 0.015 \text{ mm}$ $\text{Step 4: Find the value of } x.$ $\text{Percentage error} = \frac{\text{Mean deviation}}{\text{Mean reading}} \times 100\%$ $\Rightarrow \frac{x}{121} = \frac{0.015}{1.21} \times 100\%$ $\Rightarrow \frac{x}{121} = \frac{1.21}{150} \times 100$ $\Rightarrow \frac{x}{121} = \frac{121}{121}$ $\Rightarrow x = 150$ $\text{Hence, option (3) is the correct answer.}$

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
}