Import Question JSON

Current Question (ID: 19349)

Question:
$\text{Shown in the figure is a hollow ice cream cone (it is open at the top).}$ $\text{If its mass is } M, \text{ radius of its top, } R \text{ and height, } H, \text{ then its moment}$ $\text{of Inertia about its axis is:}$
Options:
  • 1. $\frac{M \left(R^2 + H^2\right)}{4}$
  • 2. $\frac{MR^2}{3}$
  • 3. $\frac{MR^2}{2}$
  • 4. $\frac{MH^2}{3}$
Solution:
$\text{Hint: } I = \int (dm) r^2$ $\text{Area } = \pi R l = \pi R \left(\sqrt{H^2 + R^2}\right)$ $\text{Area of element } dA = 2\pi r \, dl = 2\pi r \, \frac{dh}{\cos \theta}$ $\text{Mass of element } dm = \frac{M}{\pi R \sqrt{H^2 + R^2}} \times 2\pi r \, \frac{dh}{\cos \theta}$ $dm = \frac{2M h}{R \sqrt{H^2 + R^2}} \tan \theta \, \frac{dh}{R^2 \cos \theta}$ $\text{(here } r = h \tan \theta)$ $I = \int (dm) r^2 = \int \frac{h^2 \tan^2 \theta}{\cos \theta} \left(\frac{2m}{R} \frac{h \tan \theta}{\sqrt{R^2 + H^2}} \right) dh$ $\int_0^H h^3 dh = \frac{MR^2 H^4}{2RH^3 \sqrt{R^2 + H^2} \cos \theta} = \frac{MR^2 H \sqrt{R^2 + H^2}}{2 \sqrt{R^2 + H^2} \times H}$

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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
}