Import Question JSON

Current Question (ID: 9689)

Question:

$\text{The speed of water in a river is 4 km/h and a man can swim at 5 km/h. The minimum time taken by the man to cross the river of width 200 m is:}$

Options:

1. $\frac{1}{5} \text{ h}$
2. $\frac{1}{25} \text{ h}$
3. $\frac{1}{15} \text{ h}$
4. $\frac{1}{20} \text{ h}$

Solution:

\text{Hint: } \vec{v}_{MR} \text{ should be } \perp \text{ to } \vec{v}_{RG} \text{Given:} \text{River flow speed: } v_r = 4 \text{ km/h} \text{Man's swimming speed: } v_m = 5 \text{ km/h} \text{River width: } d = 200 \text{ m} = 0.2 \text{ km} \text{Step 1: Identify condition for minimum time} \text{For minimum time to cross a river, the swimmer should swim perpendicular to the river flow.} \text{The swimmer's velocity relative to ground should have maximum component perpendicular to the river banks.} \text{Step 2: Calculate time} \text{When swimming perpendicular to river flow, the effective crossing speed equals the swimmer's speed in still water.} \text{Time to cross} = \frac{\text{width}}{\text{swimming speed}} = \frac{0.2 \text{ km}}{5 \text{ km/h}} = \frac{1}{25} \text{ h} \text{Note: The river current affects the downstream displacement but not the crossing time when swimming perpendicular to the flow.}

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
}