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Current Question (ID: 8909)

Question:

$\text{The variation of quantity } A \text{ with quantity } B \text{ is plotted in the given figure which describes the motion of a particle in a straight line.}$ $\text{Consider the following statements:}$ $\text{(a) Quantity } B \text{ may represent time.}$ $\text{(b) Quantity } A \text{ is velocity if motion is uniform.}$ $\text{(c) Quantity } A \text{ is displacement if motion is uniform.}$ $\text{(d) Quantity } A \text{ is velocity if motion is uniformly accelerated.}$ $\text{Select the correct option:}$

Options:

1. $\text{(a), (b), (c)}$
2. $\text{(b), (c), (d)}$
3. $\text{(a), (c), (d)}$
4. $\text{(a), (c)}$

Solution:

$\text{Hint: Recall the concept of uniform motion and uniformly accelerated motion.}$ $\text{Step 1: Find the nature of the displacement-time graph in uniform motion.}$ $\text{When we are calculating the velocity of a displacement-time graph we have to take the slope. Similarly, we have to take the slope of the velocity-time graph to calculate acceleration.}$ $\text{Step 2: Find the nature of the velocity-time graph for uniformly accelerated motion.}$ $\text{When the slope is constant motion will be uniform. When we are representing motion by a graph it may be displacement-time, velocity-time or acceleration-time hence, B may represent time. For uniform motion, the velocity-time graph should be a straight line parallel to the time axis. For uniform motion velocity is constant hence slope will be positive. Hence quantity A is displacement. Therefore, for uniformly accelerated motion slope will be positive and A will represent velocity.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}