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Current Question (ID: 9775)

Question:
$\text{A block } B \text{ is placed on top of block } A. \text{ The mass of block } B \text{ is less than the mass of block } A. \text{ Friction exists between the blocks, whereas the ground on which block } A \text{ is placed is assumed to be smooth. A horizontal force } F, \text{ increasing linearly with time begins to act on } B. \text{ The acceleration } a_A \text{ and } a_B \text{ of blocks } A \text{ and } B \text{ respectively are plotted against } t. \text{ The correctly plotted graph is:}$
Options:
  • 1. $\text{Graph 1: } a_B \text{ starts above } a_A \text{ and both increase linearly with } a_B > a_A$
  • 2. $\text{Graph 2: } a_B \text{ and } a_A \text{ both start from origin and increase linearly with } a_B > a_A$
  • 3. $\text{Graph 3: } a_B \text{ starts below } a_A \text{ and both increase with } a_B < a_A \text{ initially}$
  • 4. $\text{Graph 4: } a_B \text{ and } a_A \text{ both start from origin, } a_A \text{ remains constant while } a_B \text{ increases}$
Solution:
\text{Analysis of the problem:} \text{Initially, when the applied force is less than limiting friction between block A and B, the whole system moves with common acceleration:} a_A = a_B = \frac{F}{m_A + m_B} \text{But as the applied force increases with time, block B starts moving under the effect of net force } F - F_k \text{Where } F_k = \text{kinetic friction between block A and B} \text{Acceleration of block B: } a_B = \frac{F - F_k}{m_B} \text{As F is increasing with time, so } a_B \text{ will increase with time} \text{Kinetic friction is the cause of motion of block A} \text{Acceleration of block A: } a_A = \frac{F_k}{m_A} \text{It is clear that } a_B > a_A \text{, i.e. graph (4) correctly represents the variation in acceleration with time for block A and B.}

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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
}