Import Question JSON

Current Question (ID: 10329)

Question:

$\text{A circular disc is to be made by using iron and aluminium so that it acquires a maximum moment of inertia about its geometrical axis. It is possible with:}$

Options:

1. $\text{Aluminium in the interior and iron surrounding it}$
2. $\text{Iron at the interior and aluminium surrounding it}$
3. $\text{Using iron and aluminium layers in alternate order}$
4. $\text{A sheet of iron is used at both the external surface and aluminium sheet as the internal layer}$

Solution:

$\text{Complete step-by-step answer:}$ $\text{A circular disc is made up of a larger number of circular rings. Total moment of inertia of disc can be given by the sum of moment of inertia of all these small rings about the geometrical axis. Now the moment of inertia of a circular ring is given by-}$ $I = MR^2$ $\Rightarrow I \propto M$ $\text{Since mass is proportional to the density of the material. As we know the density of the Iron is much more than that of aluminium. Hence to get maximum value of moment of Inertia the less dense material should be used at interior and denser at the surroundings of it. Therefore, using Aluminium at the interior and Iron at its surrounding will maximizes the moment of inertia.}$ $\text{Hence option (a) is the correct answer.}$ $\text{If we use Iron and aluminium layers in alternate order then it will increase the mass but not maximise it. Hence option (a) is correct.}$ $\text{Iron at the interior and aluminium surrounding it will lead to the minimization of mass. Hence option (c) is also correct.}$ $\text{We can't use both at the same time, so option (d) is not correct.}$ $\text{The correct option is (a)}$

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
}