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Current Question (ID: 10234)

Question:

$\text{If the linear density of a rod of length 3 m varies as } \lambda = 2 + x\text{, then the position of the center of mass of the rod is at a distance of:}$

Options:

1. $\frac{7}{3} \text{ m}$
2. $\frac{10}{7} \text{ m}$
3. $\frac{12}{7} \text{ m}$
4. $\frac{9}{7} \text{ m}$

Solution:

\text{Hint: } x_{\text{com}} = \frac{\int x \, dm}{\int dm} \text{Step 1: Find the position of the center of mass of the rod.} \text{The centre of mass is given by } x_{\text{com}} = \frac{1}{M} \int_0^L x \lambda(x) \, dx \text{Step 2: Calculate the total mass of the rod which is given by} M = \int_0^L \lambda(x) \, dx M = \int_0^3 (2 + x) \, dx = \left[2x + \frac{x^2}{2}\right]_0^3 M = 6 + \frac{9}{2} = 10.5 \text{ kg} \text{Step 3: Calculate the center of mass of the rod is given by} x_{\text{com}} = \frac{1}{M} \int_0^L x \lambda(x) \, dx x_{\text{com}} = \frac{1}{10.5} \int_0^3 x(2 + x) \, dx = \frac{1}{10.5} \left[x^2 + \frac{x^3}{3}\right]_0^3 x_{\text{com}} = \frac{1}{10.5} \left[9 + \frac{27}{3}\right] = \frac{1}{10.5} \times 18 = \frac{18}{10.5} = \frac{12}{7} \text{ m} \text{The position of the center of mass of the rod is at a distance of } \frac{12}{7} \text{ m from the origin.} \text{Hence, option (3) is the correct answer.}

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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
}