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Current Question (ID: 19018)

Question:

$\text{Two stones are thrown simultaneously from the edge of a cliff 240 m high.}$ $\text{The first stone is thrown upward with an initial speed of 10 m/s, and the second with 40 m/s.}$ $\text{Assuming the stones do not bounce after hitting the ground and neglecting air resistance } (g = 10 \text{ m/s}^2),$ $\text{which of the following graphs best represents how the position of the second stone varies relative to the first stone with time?}$ $\text{(graphs are schematic and not drawn to scale)}$

Options:

1. $(y_2 - y_1) \text{ m}$ $t \text{ (s)}$
2. $(y_2 - y_1) \text{ m}$ $t \text{ (s)}$
3. $(y_2 - y_1) \text{ m}$ $t \text{ (s)}$
4. $(y_2 - y_1) \text{ m}$ $t \text{ (s)}$

Solution:

$\text{Hint: } s = ut + \frac{1}{2} at^2$ $\text{(i) Time taken by A to reach the ground}$ $5t^2 - 10t - 240 = 0$ $t = 8 \text{ s}$ $\text{and time taken by B to reach the ground}$ $-240 = 40t - 5t^2$ $t = 12 \text{ s}$ $\text{(ii) For } 0 \leq t \leq 8 \text{ sec}$ $y_1 = 10t - 5t^2$ $y_2 = 40t - 5t^2$ $y_2 - y_1 = 40t - 10t$ $y_2 - y_1 = 30t$ $\text{For } 0 \leq t \leq 8 \text{ graph will be straight line}$ $\text{(iii) at } t = 8 \text{ sec,}$ $y_1 = -240 \text{ m}$ $y_2 = 40(8) - 5(8)^2 = 0$ $\text{(iv) for } 8 \leq t \leq 12$ $y_1 = -240$ $y_2 = 40t - 5t^2$ $y_2 - y_1 = 40t - 5t^2 + 240$ $y_2 - y_1 = -5t^2 + 40t + 240$ $\text{So graph will be parabolic with -ve slope}$

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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
}