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Current Question (ID: 9765)

Question:

$\text{On the horizontal surface of a truck, a block of mass 1 kg is placed (} \mu = 0.6 \text{) and the truck is moving with an acceleration of } 5 \text{ m/s}^2\text{. The frictional force on the block will be:}$

Options:

1. $5 \text{ N}$
2. $6 \text{ N}$
3. $5.88 \text{ N}$
4. $8 \text{ N}$

Solution:

\text{Given:} \text{Mass of block, } m = 1 \text{ kg} \text{Coefficient of friction, } \mu = 0.6 \text{Acceleration of truck, } a = 5 \text{ m/s}^2 \text{Analysis:} \text{Frictional force, } f = \mu_s N = \mu_s mg = 0.6 \times 1 \times 10 = 6 \text{ N} \text{Pseudo force, } F = ma = 1 \times 5 = 5 \text{ N} \text{Here, } F < f \text{ so the block does not move relative to the truck.} \text{Therefore, applied force = frictional force = 5 N}

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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
}