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Current Question (ID: 19350)

Question:

$\text{The linear mass density of a thin rod AB of length L varies from A to B as } \lambda(x) = \lambda_0 \left(1 + \frac{x}{L}\right), \text{ where } x \text{ is the distance from A. If the mass } M \text{ of the rod is } \frac{3 \lambda_0 L}{2}, \text{ then its moment of inertia about an axis passing through A and perpendicular to the rod is:}$

Options:

1. $\frac{5}{12} ML^2$
2. $\frac{3}{7} ML^2$
3. $\frac{2}{5} ML^2$
4. $\frac{7}{18} ML^2$

Solution:

$\text{Hint: } MOI \text{ is given by, } I = \int dmx^2$ $\text{Step 1: Find the MOI of the rod.}$ $\text{Here, } dm = \lambda dx$ $dI = dmx^2$ $I = \int dI = \int dmx^2 = \int \lambda dx \times x^2 = \int_0^x \lambda_0 \left[1 + \frac{x}{L}\right] x^2 dx$ $I = \lambda_0 \left[\frac{x^3}{3} + \frac{x^4}{4L}\right]_0^L = \lambda_0 \left[\frac{L^3}{3} + \frac{L^3}{4}\right] = \lambda_0 \times \frac{7}{12} L^3$ $\text{Step 2: Express MOI in terms of M.}$ $I = \lambda_0 \times \frac{7}{12} \times L^3 \times \frac{M}{M} = \lambda_0 \times \frac{7}{12} L^3 \times \frac{M}{\frac{3 \lambda_0 L}{2}}$ $= \frac{7}{18} ML^2$

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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
}