Import Question JSON

Current Question (ID: 9592)

Question:
$\text{A man sitting in a bus travelling in a direction from west to east with a speed of 40 km/h observes that the rain-drops are falling vertically downwards. To another man standing on ground the rain will appear:}$
Options:
  • 1. $\text{To fall vertically downwards}$
  • 2. $\text{To fall at an angle going from west to east}$
  • 3. $\text{To fall at an angle going from east to west}$
  • 4. $\text{The information given is insufficient to decide the direction of the rain}$
Solution:
$\text{A man is sitting in a bus and travelling from west to east, and the rain drops are appears falling vertically down.}$ $\text{Let us define the variables:}$ $v_m = \text{velocity of man}$ $v_r = \text{Actual velocity of rain which is falling at an angle } \theta \text{ with vertical}$ $v_{rm} = \text{velocity of rain w.r.t. to moving man}$ $\text{From the vector diagram, we can see that the man in the bus observes rain falling vertically because the horizontal component of rain velocity equals the bus velocity.}$ $\text{However, if another man observes the rain from the ground, he will find that rain is actually falling with velocity } v_r \text{ at an angle going from west to east.}$ $\text{This is because the stationary observer sees the true motion of raindrops, which includes both vertical and horizontal components.}$ $\text{The horizontal component of rain velocity is in the same direction as the bus motion (west to east).}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}