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Current Question (ID: 10135)

Question:

$\text{A light rod of length } l \text{ has two masses, } m_1 \text{ and } m_2, \text{ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:}$

Options:

1. $\frac{m_1m_2}{m_1+m_2}l^2$
2. $\frac{m_1+m_2}{m_1m_2}l^2$
3. $(m_1 + m_2)l^2$
4. $\sqrt{(m_1m_2)}l^2$

Solution:

$\text{The moment of inertia of the point masses about the given axis is,}$ $\Rightarrow I = m_1x_1^2 + m_2x_2^2$ $\text{Where } x_1 \text{ and } x_2 \text{ are the distances of masses } m_1 \text{ and } m_2 \text{ from the center of mass C.}$ $\text{For center of mass: } x_1 = \frac{m_2l}{m_1+m_2} \text{ and } x_2 = \frac{m_1l}{m_1+m_2}$ $= m_1\left(\frac{m_2l}{m_1+m_2}\right)^2 + m_2\left(\frac{m_1l}{m_1+m_2}\right)^2$ $= \frac{m_1m_2l^2}{(m_1+m_2)^2}(m_2 + m_1)$ $= \frac{m_1m_2l^2}{(m_1+m_2)}$

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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
}