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Current Question (ID: 19196)

Question:

$\text{A block of mass } 40 \text{ kg slides over a surface when a mass of } 4 \text{ kg is suspended through an inextensible massless string passing over a frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is } 0.02. \text{ The acceleration of the block is: (Given } g = 10 \text{ ms}^{-2})$

Options:

1. $1 \text{ ms}^{-2}$
2. $\frac{1}{5} \text{ ms}^{-2}$
3. $\frac{4}{5} \text{ ms}^{-2}$
4. $\frac{8}{11} \text{ ms}^{-2}$

Solution:

$\text{Let the tension in the string be } T. \text{ For the } 4 \text{ kg mass: } 4g - T = 4a.$ $\text{For the } 40 \text{ kg block: } T - \mu \times 40g = 40a.$ $\text{Solving these equations: } 4g - 40 \mu g = 44a.$ $a = \frac{4g - 40 \mu g}{44} = \frac{4 \times 10 - 40 \times 0.02 \times 10}{44} = \frac{8}{11} \text{ ms}^{-2}.$

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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
}