Import Question JSON

Current Question (ID: 7340)

Question:

\text{Two students performed the same experiment separately, and each of them recorded two readings of mass, which are given below. The correct reading of mass is } 3.0 \text{ g.} \text{Student A: Reading 1 = } 3.01, \text{ Reading 2 = } 2.99 \text{Student B: Reading 1 = } 3.05, \text{ Reading 2 = } 2.95 \text{On the basis of the given data, mark the correct option out of the following statements.}

Options:

1. $\text{The results of both the students' are neither accurate nor precise.}$
2. $\text{The results of student A are both precise and accurate.}$
3. $\text{Student B's results are neither precise nor accurate.}$
4. $\text{The results of student B are both precise and accurate.}$

Solution:

\text{Hint: If the calculated values are close to each other then calculated data is precise} \text{The average value of A and B are as follows:} \text{The average value of A} = \frac{3.01 + 2.99}{2} = 3 \text{The average value of B} = \frac{3.05 + 2.95}{2} = 3 \text{Both A and B average value is close to the accurate value. Hence, both values are accurate.} \text{The gap between values of A and B is not much, hence, the values are precise.} \text{Therefore, option second 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
}