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Current Question (ID: 19385)

Question:
$\text{Two solid spheres of mass } m = \frac{1}{2} \text{ kg each are connected at the ends of a light rod as shown in the figure.}$ $\text{The assembly rotates about axis AA'. Then moment of inertia of the assembly is equal to } \frac{x}{5} \text{ kgm}^2 \text{ value of } x \text{ is equal to:}$
Options:
  • 1. 1.27
  • 2. 3.14
  • 3. 2.20
  • 4. 2.91
Solution:
$\text{Hint: Use parallel axes theorem.}$ $MI = \left[ \left( \frac{2}{5} Mr^2 \right) + (MR^2) \right] \times 2$ $= \left[ \frac{2}{5} \times \frac{1}{2} \times (0.1)^2 + \frac{1}{2} \times (0.5)^2 \right] \times 2$ $= \frac{0.02}{5} + \frac{1.25}{2}$

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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
}