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Current Question (ID: 10110)

Question:
$\text{The mass per unit length of a non-uniform rod of length } L \text{ is given by } \mu = \lambda x^2 \text{ where } \lambda \text{ is a constant and } x \text{ is the distance from one end of the rod. The distance between the centre of mass of the rod and this end is:}$
Options:
  • 1. $\frac{L}{2}$
  • 2. $\frac{L}{4}$
  • 3. $\frac{3L}{4}$
  • 4. $\frac{L}{3}$
Solution:
$\text{Consider an element of length } dx \text{ at a distance } x.$ $\text{mass of the element} = \lambda x^2 dx$ $x_{\text{cm}} = \frac{\int x dm}{\int dm} = \frac{\int_0^L \lambda x^3 dx}{\int_0^L \lambda x^2 dx} = \frac{\left[\frac{x^4}{4}\right]_0^L}{\left[\frac{x^3}{3}\right]_0^L}$ $x_{\text{cm}} = \frac{3L}{4}$

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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
}