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Current Question (ID: 19315)

Question:
$\text{Distance of the centre of mass of a solid uniform cone from its vertex is } Z_0. \text{ If the radius of its base is } R \text{ and its height is } h \text{ then } Z_0 \text{ is equal to:}$
Options:
  • 1. $\frac{h^2}{4R}$
  • 2. $\frac{3h}{4}$
  • 3. $\frac{4}{5h}$
  • 4. $\frac{3h^2}{8R}$
Solution:
$\text{Hint: } X_c = \frac{\int x \, dm}{M}$ $\therefore \frac{r}{x} = \frac{R}{h}$ $r = \frac{R}{h} x$ $z_0 = \text{distance of CM from O} = x_c = \frac{\int dm \, x}{M} = \frac{\int \rho \, (\pi r^2 \, dx) \, x}{M}$ $x_c = \frac{\pi \rho \int \left( \frac{R}{h} x \right)^2 \, dx \, x}{M}$ $x_c = \pi \cdot \frac{M}{\frac{1}{3} \pi R^2 h} \cdot \frac{R^2}{h^2} \int_0^h x^3 \, dx$ $z_0 = x_c = \frac{3}{h^3} \cdot \frac{h^4}{4} = \frac{3h}{4}$

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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
}