Import Question JSON

Current Question (ID: 10119)

Question:

$\text{If a rod of length 3 m with its mass acting per unit length, is directly proportional to distance } x \text{ from one of its ends, then its centre of gravity from that end will be at:}$

Options:

1. $\text{1.5 m}$
2. $\text{2 m}$
3. $\text{2.5 m}$
4. $\text{3.0 m}$

Solution:

$\text{Here } \rho = kx \text{ where } k \text{ is a constant mass of small element of } dx \text{ length is}$ $dm = kx \cdot dx$ $x_{\text{cm}} = \frac{\int x \cdot dm}{\int dm} = \frac{\int_0^3 x \left( x \right) dx}{\int_0^3 x \cdot dx} = \frac{\left[ \frac{x^3}{3} \right]_0^3}{\left[ \frac{x^2}{2} \right]_0^3} = \frac{\frac{27}{3}}{\frac{9}{2}} = 2$

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
}