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Current Question (ID: 19068)

Question:

$\text{A train (moving with initial speed } = 20 \text{ m/s) applies brakes to stop at the incoming station which is } 500 \text{ m ahead. If brakes are applied after moving } 250 \text{ m, then how much beyond the station train would stop?}$

Options:

1. $125 \text{ m}$
2. $500 \text{ m}$
3. $250 \text{ m}$
4. $400 \text{ m}$

Solution:

$\text{Hint: Use the concept of stopping distance.}$ $\text{Step: Find the displacement beyond the station where the train will stop.}$ $\text{With the help of stopping distance, we can calculate the retardation of the train.}$ $u = 20 \text{ m/s}, s = 500 \text{ m}, v = 0 \text{ m/s}$ $\text{Apply the kinematic equation of motion}$ \Rightarrow v^2 = u^2 + 2as \Rightarrow 0 = 20^2 + 2a(500) \Rightarrow a = -0.4 \text{ m/s}^2$ $\text{The train needs } 500 \text{ m to stop.}$ $\text{So, it will move beyond the station by } \Rightarrow 500 - 250 = 250 \text{ m.}$ $\text{Option (3) is correct.}$

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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
}