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Current Question (ID: 8891)

Question:

$\text{A particle is moving along the x-axis such that its velocity varies with time as per the equation } v = 20\left(1 - \frac{t}{2}\right)\text{. At } t = 0 \text{ particle is at the origin. From the following, select the correct position }(x)\text{ - time }(t)\text{ plot for the particle:}$

Options:

1. $\text{Graph 1: Linear decrease in position with time, reaching a minimum and then continuing downward}$
2. $\text{Graph 2: Position decreases linearly, reaches a minimum at t = 2s, then increases linearly}$
3. $\text{Graph 3: Position decreases in a curved manner, reaching a minimum at t = 2s, then increases}$
4. $\text{Graph 4: Parabolic curve starting from origin, reaching maximum at t = 2s, then returning to origin at t = 4s}$

Solution:

$\text{Hint: } v = \frac{dx}{dt}$ $\text{Step: Find the expression for x.}$ $\text{The displacement of the particle is given by;}$ $\Rightarrow \int_0^x dx = \int_0^t v\,dt = \int_0^t (20 - 10t)\,dt$ $\text{Integrating: } x = 20t - 5t^2 \quad \text{...(1)}$ $\text{Therefore, the above expression (1) represents an equation of a parabola as shown in figure below;}$ $\text{Analysis of the motion:}$ $\text{At } t = 0: x = 0 \text{ (starts at origin)}$ $\text{At } t = 2\text{s: } x = 20(2) - 5(4) = 20 \text{ (maximum displacement)}$ $\text{At } t = 4\text{s: } x = 20(4) - 5(16) = 0 \text{ (returns to origin)}$ $\text{The velocity becomes zero at } t = 2\text{s, and negative afterwards, causing the particle to return.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}