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Current Question (ID: 19368)

Question:
$\text{A uniform disc with mass } M = 4 \text{ kg and radius } R = 10 \text{ cm is mounted on a fixed horizontal axle as shown in the figure. A block with mass } m = 2 \text{ kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is: (Take } g = 10 \text{ ms}^{-2})$
Options:
  • 1. $12 \text{ N}$
  • 2. $20 \text{ N}$
  • 3. $10 \text{ N}$
  • 4. $15 \text{ N}$
Solution:
$\text{Hint: } \tau = I \alpha$ $\text{Step: Find the tension in the string.}$ $a = R \alpha \ldots (i)$ $mg - T = ma \ldots (ii)$ $TR = I \alpha \ldots (iii)$ $mg = T + m(R \alpha)$ $= T + m\left(\frac{R \times 2TR}{MR^2}\right)$ $= 2T$ $\Rightarrow T = \frac{mg}{2} = 10 \text{ N}$

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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
}