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Current Question (ID: 10493)

Question:
$\text{A small sphere of radius } r \text{ falls from rest in a viscous liquid. As a result, heat is produced due to the viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:}$
Options:
  • 1. $r^3$
  • 2. $r^2$
  • 3. $r^5$
  • 4. $r^4$
Solution:
\text{If the ball falls in a viscous liquid, then at terminal velocity the viscous force balances the gravitational force.} \frac{dQ}{dt} = F_v \cdot v \frac{dQ}{dt} = 6\pi\eta rv^2 \text{As we know, } v = \frac{2r^2(\sigma - \rho)g}{9\eta} \text{Substituting this into the heat production equation:} \frac{dQ}{dt} \propto r \cdot r^4 \frac{dQ}{dt} \propto r^5 \text{Therefore, the rate of heat production is proportional to } r^5\text{.}

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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
}