Import Question JSON

Current Question (ID: 19176)

Question:

$\text{The acceleration of the 2 kg block is shown in the diagram is:}$ $\text{(neglect friction)}$

Options:

1. $\frac{4g}{15}$
2. $\frac{2g}{15}$
3. $\frac{15}{g}$
4. $\frac{2g}{3}$

Solution:

$\text{Hint: } F = ma$ $\text{For 2 kg block}$ $T - 2g \sin 37^\circ = 2a \quad \text{(i)}$ $\text{For 4 kg block}$ $4g - 2T = \frac{4a}{2}$ $2g - T = a \quad \text{(ii)}$ $T = (2g - a)$ $3a = 2g \times \frac{2}{5}$ $2g - a - 2g \times \frac{3}{5} = 2a$ $a = \frac{4g}{15}$

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
}