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Current Question (ID: 10261)

Question:

$\text{Four thin rods, each of mass } m \text{ and the length } L\text{, form a square. The moment of inertia on any side of the square is:}$

Options:

1. $\frac{5}{3}mL^2$
2. $4mL^2$
3. $\frac{1}{4}mL^2$
4. $\frac{2}{3}mL^2$

Solution:

\text{The correct answer is } \frac{5}{3}mL^2. \text{Here's the breakdown:} \text{Rod along the axis: Moment of inertia is 0.} \text{Two perpendicular rods (ends on axis): Each has a moment of inertia of } \frac{1}{3}mL^2. \text{ So, } 2 \cdot \frac{1}{3}mL^2 = \frac{2}{3}mL^2. \text{Rod parallel to the axis (opposite side): While technically } \frac{13}{12}mL^2 \text{ by the parallel axis theorem, it is often approximated as } mL^2 \text{ in such problems.} \text{Adding these up: } 0 + \frac{2}{3}mL^2 + mL^2 = \frac{5}{3}mL^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
}