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Current Question (ID: 10363)

Question:

A steel ring of radius $r$ and cross-section area $A$ is fitted onto a wooden disc of radius $R(R > r)$. If Young's modulus is $E$, then the force with which the steel ring is expanded is:

Options:

1. $AE\frac{R}{r}$
2. $AE\left(\frac{R-r}{r}\right)$
3. $\frac{E}{A}\left(\frac{R-r}{A}\right)$
4. $\frac{Er}{AR}$

Solution:

Initial length (circumference) of the ring $= 2\pi r$ Final length (circumference) of the ring $= 2\pi R$ Change in length $= 2\pi R - 2\pi r$ $\text{strain} = \frac{\text{change in length}}{\text{original length}} = \frac{2\pi(R-r)}{2\pi r} = \frac{R-r}{r}$ Now, Young's modulus $E = \frac{F/A}{l/L} = \frac{F/A}{(R-r)/r}$ $\therefore F = AE\left(\frac{R-r}{r}\right)$

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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
}