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Current Question (ID: 19964)

Question:
$\text{A uniform string of length } 20 \text{ m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take } g = 10 \text{ ms}^{-2})$
Options:
  • 1. $2\pi \sqrt{2} \text{ s}$
  • 2. $2 \text{ s}$
  • 3. $2\sqrt{2} \text{ s}$
  • 4. $\sqrt{2} \text{ s}$
Solution:
$\text{Hint: } v = \sqrt{\frac{T}{\mu}}$ $\text{Let } m \rightarrow \text{ mass/length}$ $T = (xm)g$ $V = \sqrt{\frac{T}{m}}$ $= \sqrt{\frac{(xm)g}{m}}$ $= \sqrt{gx}$ $a = V \frac{dV}{dx} = \sqrt{gx} \frac{d}{dx} \sqrt{gx}$ $= \sqrt{gx} \times \sqrt{g} \times \frac{1}{2\sqrt{x}}$ $a = \frac{g}{2} = \text{ constant}$ $\therefore l = \frac{1}{2}at^2$ $t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l}{g/2}} = 2\sqrt{\frac{20}{20}} = 2\sqrt{2} \text{ s}$

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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
}