Import Question JSON

Current Question (ID: 9736)

Question:
$\text{Three blocks A, B, and C of masses } 4 \text{ kg, } 2 \text{ kg, and } 1 \text{ kg respectively, are in contact on a frictionless surface, as shown. If a force of } 14 \text{ N is applied to the } 4 \text{ kg block, then the contact force between A and B is:}$
Options:
  • 1. $2 \text{ N}$
  • 2. $6 \text{ N}$
  • 3. $8 \text{ N}$
  • 4. $18 \text{ N}$
Solution:
$\text{Hint: All the blocks will have the same acceleration.}$ $\text{Step 1: Find the common acceleration of the blocks.}$ $\text{Net mass of the system, } M = 4 + 2 + 1 = 7 \text{ kg}$ $\text{Force } F = 14 \text{ N}$ $\text{As } F = Ma\text{,}$ $a = \frac{F}{M} = \frac{14}{7} = 2 \text{ m/s}^2$ $\text{Step 2: Apply Newton's second law for block A.}$ $\text{FBD of block A:}$ $\text{Applied force } F = 14 \text{ N (rightward)}$ $\text{Contact force from B on A } = F' \text{ (leftward)}$ $F - F' = 4a$ $F' = F - 4a = 14 - 4 \times 2 = 14 - 8 = 6 \text{ N}$

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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
}