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Current Question (ID: 10488)

Question:

$\text{A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, then the distance between the plates in m is:}$

Options:

1. $0.1$
2. $0.05$
3. $0.005$
4. $0.0005$

Solution:

\text{Given:} \text{Area } A = (0.1)^2 = 0.01 \text{ m}^2 \text{Coefficient of viscosity } \eta = 0.01 \text{ Poise} = 0.001 \text{ decapoise (M.K.S. unit)} \text{Velocity } dv = 0.1 \text{ m/s and Force } F = 0.002 \text{ N} \text{Using the formula for viscous force: } F = \eta A \frac{dv}{dx} \text{Rearranging to find distance: } dx = \frac{\eta A \cdot dv}{F} = \frac{0.001 \times 0.01 \times 0.1}{0.002} = 0.0005 \text{ m}

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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
}