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Current Question (ID: 9795)

Question:

$\text{A ball of mass } 0.1 \text{ kg is whirled in a horizontal circle of radius } 1 \text{ m by means of a string at an initial speed of } 10 \text{ rpm. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is:}$

Options:

1. $5 \text{ rpm}$
2. $10 \text{ rpm}$
3. $20 \text{ rpm}$
4. $14 \text{ rpm}$

Solution:

\text{Given the relationship: } T \sin \theta = m\omega^2 r \text{Since } \sin \theta \text{ is constant, } T \propto \omega^2 \text{, therefore } \omega \propto \sqrt{T} \text{When tension is reduced to one quarter: } \frac{\omega_2}{\omega_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{T/4}{T}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \text{Therefore: } \omega_2 = \frac{\omega_1}{2} = \frac{10}{2} = 5 \text{ rpm}

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}