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Current Question (ID: 16899)

Question:

$\text{A thin rectangular magnet suspended freely has a period of oscillation equal to } T. \text{ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is } T', \text{ then ratio } \frac{T'}{T} \text{ is:}$

Options:

1. $\frac{1}{4}$
2. $\frac{1}{2\sqrt{2}}$
3. $\frac{1}{2}$
4. $2$

Solution:

$T = 2\pi \sqrt{\frac{I}{MB}}$ $\frac{T_1}{T_2} = \sqrt{\frac{I_1M_2}{I_2M_1}} = \sqrt{\frac{I}{\frac{I}{8}} \times \frac{\frac{M}{2}}{M}} = 2$ $\left[ I_2 = \frac{M}{2} \times \frac{(\frac{L}{2})^2}{12} = \frac{1}{8} \left( \frac{ML^2}{12} \right) = \frac{I}{8} \right]$ $\frac{T_2}{T_1} = \frac{1}{2}$

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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
}