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Current Question (ID: 10102)

Question:

$\text{A uniform square plate } ABCD \text{ has a mass of 10 kg. If two point masses of 5 kg each are placed at the corners } C \text{ and } D \text{ as shown in the adjoining figure, then the centre of mass shifts to the mid-point of:}$

Options:

1. $OH$
2. $DH$
3. $OG$
4. $OF$

Solution:

$\text{Hint: Treat the square plate as a point.}$ $\text{Step 1: Locate the centre of mass of the square plate.}$ $\text{Suppose the centre of the square is located at the origin.}$ $\text{The square plate (10 kg) is at origin O, and the two 5 kg masses are at corners C and D.}$ $\text{Setting up coordinates with O at origin, if the square has side length } 2a\text{:}$ $\text{- Square plate: mass = 10 kg at }(0, 0)$ $\text{- Point mass at C: mass = 5 kg at }(a, -a)$ $\text{- Point mass at D: mass = 5 kg at }(-a, -a)$ $\text{Step 2: Find the centre of mass for the system of point masses.}$ $x_{cm} = \frac{10(0) + 5(a) + 5(-a)}{10 + 5 + 5} = \frac{0}{20} = 0$ $y_{cm} = \frac{10(0) + 5(-a) + 5(-a)}{20} = \frac{-10a}{20} = -\frac{a}{2}$ $\text{Therefore, the new center of mass is at }(0, -\frac{a}{2})\text{, which is the midpoint of } OH\text{.}$

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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
}