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Current Question (ID: 19193)

Question:

$\text{A system of two blocks of masses } m = 2 \text{ kg and } M = 8 \text{ kg is placed on a smooth table as shown in the figure.}$ $\text{The coefficient of static friction between two blocks is } 0.5. \text{ The maximum horizontal force } F$ $\text{that can be applied to the block of mass } M \text{ so that the blocks move together will be:}$

Options:

1. $9.8 \text{ N}$
2. $39.2 \text{ N}$
3. $49 \text{ N}$
4. $78.4 \text{ N}$

Solution:

$\text{The maximum static friction force } f_s = \mu_s \cdot m \cdot g = 0.5 \cdot 2 \cdot 9.8 = 9.8 \text{ N}$ $\text{This force is the maximum force that can be applied to } m \text{ without slipping.}$ $\text{For the blocks to move together, } F = f_s + M \cdot a = 9.8 + 8 \cdot a$ $\text{Since } a = \frac{f_s}{m} = \frac{9.8}{2} = 4.9 \text{ m/s}^2$ $F = 9.8 + 8 \cdot 4.9 = 49 \text{ N}$

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
}