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Current Question (ID: 10256)

Question:

$\text{In the three figures, each wire has a mass M, radius R and a uniform mass distribution. If they form part of a circle of radius R, then about an axis perpendicular to the plane and passing through the centre (shown by crosses), their moment of inertia is in the order:}$

Options:

1. $I_A > I_B > I_C$
2. $I_A = I_B = I_C$
3. $I_A < I_B < I_C$
4. $I_A < I_C < I_B$

Solution:

$\text{The moment of inertia (I) of a rigid body about an axis is given by the integral } I = \int r^2 dm\text{, where } r \text{ is the perpendicular distance from the mass element } dm \text{ to the axis of rotation.}$ $\text{In this problem, we have wires forming parts of a circle, each with mass M and radius R. The axis of rotation passes through the center of the circle (indicated by crosses) and is perpendicular to the plane of the figures. For all points on the wire, the distance } r \text{ from the axis of rotation is constant and equal to R.}$ $\text{Therefore, for each wire, the moment of inertia can be calculated as:}$ $I = \int R^2 dm = R^2 \int dm = R^2 M$ $\text{Since all three figures (A, B, and C) have the same mass M and the same radius R, their moments of inertia about the specified axis will be identical.}$ $\text{So, } I_A = I_B = I_C\text{.}$ $\text{The final answer is } 2\text{.}$

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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
}