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Current Question (ID: 19341)

Question:

$ABC$ \text{ is a plane lamina of the shape of an equilateral triangle. } D, E \text{ are mid points of } AB, AC \text{ and } G \text{ is the centroid of the lamina. The moment of inertia of the lamina about an axis passing through } G \text{ and perpendicular to the plane } ABC \text{ is } I_0. \text{ If part } ADE \text{ is removed, the moment of inertia of the remaining part about the same axis is } \frac{NI_0}{16} \text{ where } N \text{ is an integer. Value of } N \text{ is:}$

Options:

1. $3$
2. $20$
3. $11$
4. $38$

Solution:

$\text{Hint: } I = \int r^2 dm$ $\text{Let side of triangle is } a \text{ and mass is } m$ $\text{MOI of plate } ABC \text{ about centroid}$ $I_0 = \frac{m}{3} \left( \left( \frac{a}{2\sqrt{3}} \right)^2 \times 3 \right) = \frac{ma^2}{12}$ $\text{triangle } ADE \text{ is also an equilateral triangle of side } a/2.$ $\text{Let moment of inertia of triangular plant } ADE \text{ about it's centroid (G') is } I_1 \text{ and mass is } m_1.$ $m_1 = \frac{m}{\sqrt{3}a^2} \times \frac{\sqrt{4}}{4} \left( \frac{a}{2} \right)^2 = \frac{m}{4}$ $I_1 = \frac{m_1}{12} \left( \frac{a}{2} \right)^2 = \frac{m}{4 \times 12} \frac{a^2}{4} = \frac{ma^2}{192}$ $\text{distance } GG' = \frac{a}{\sqrt{3}} - \frac{a}{2\sqrt{3}} = \frac{a}{2\sqrt{3}}$ $\text{so MOI of part } ADE \text{ about centroid}$ $I_2 = I_1 + m_1 \left( \frac{a}{2\sqrt{3}} \right)^2 = \frac{ma^2}{192} + \frac{m}{4} \cdot \frac{a^2}{12}$ $= \frac{5ma^2}{192}$ $\text{now MOI of remaining part}$ $= \frac{ma^2}{12} - \frac{5ma^2}{192} = \frac{11ma^2}{12 \times 16} = \frac{11I_0}{16}$ $\Rightarrow N = 11$

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{
  "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
}