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Current Question (ID: 10196)

Question:

$\text{A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound around the cylinder with one end attached to it and the other end hanging freely. The tension in the string required to produce an angular acceleration of 2 rev/s}^2 \text{ will be:}$

Options:

1. $25 \text{ N}$
2. $50 \text{ N}$
3. $78.5 \text{ N}$
4. $157 \text{ N}$

Solution:

$\text{Hint: } \tau = I\alpha$ $\text{Step: Find the tension in the string.}$ $M = 50 \text{ kg}$ $R = 0.5 \text{ m}$ $\alpha = 2 \text{ rev/s}^2$ $\text{Torque, } \tau = TR$ $I = \frac{1}{2}MR^2$ $\text{Angular acceleration of the cylinder } \alpha = \frac{\tau}{I} = \frac{TR}{\frac{1}{2}MR^2}$ $\Rightarrow \frac{MR\alpha}{2} = \frac{50 \times 0.5 \times 4\pi}{2}$ $= 157 \text{ N}$ $\text{Hence, option (4) is the correct answer.}$

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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
}