Import Question JSON

Current Question (ID: 10487)

Question:

$\text{A metal block of area 0.10 m}^2 \text{ is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in the figure below. A liquid film with a thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m/s. The coefficient of viscosity of the liquid is:}$

Options:

1. $4.45 \times 10^{-2} \text{ Pa/s}$
2. $4.45 \times 10^{-3} \text{ Pa/s}$
3. $3.45 \times 10^{-2} \text{ Pa/s}$
4. $3.45 \times 10^{-3} \text{ Pa/s}$

Solution:

\text{Hint: Shear stress on the fluid = F/A.} \text{Step 1: Find the tension in the string.} \text{The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m.} F = T = mg = 0.01 \text{ kg} \times 9.8 \text{ m s}^{-2} = 9.8 \times 10^{-2} \text{ N} \text{Step 2: Find the coefficient of viscosity of the liquid.} \text{Shear stress on the fluid} = \frac{F}{A} = \frac{9.8 \times 10^{-2}}{0.1} = 0.98 \text{ N m}^{-2} \text{Strain rate} = \frac{v}{t} = \frac{0.085}{0.30 \times 10^{-3}} = 283.33 \text{ s}^{-1} \eta = \frac{\text{stress}}{\text{strain rate}} \eta = \frac{0.98 \text{ N m}^{-2}}{283.33 \text{ s}^{-1}} = 3.46 \times 10^{-3} \text{ Pa·s}

Import JSON File

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
}