Import Question JSON

Current Question (ID: 9672)

Question:

$\text{A man is walking on a horizontal road at a speed of 4 km/hr. Suddenly, the rain starts vertically downwards with a speed of 7 km/hr. The magnitude of the relative velocity of the rain with respect to the man is:}$

Options:

1. $\sqrt{33} \text{ km/hr}$
2. $\sqrt{65} \text{ km/hr}$
3. $8 \text{ km/hr}$
4. $4 \text{ km/hr}$

Solution:

\text{Hint: } v_{\text{relative}} = \sqrt{v_r^2 + v_m^2} \text{Step: Find the magnitude of the relative velocity of the rain with respect to the man.} \text{For the relative velocity of the rain with respect to the man, we use the concept of vector addition.} \text{The velocity of the man: } \vec{v}_m = 4\hat{i} \text{ km/hr (along the horizontal direction)} \text{The velocity of the rain: } \vec{v}_r = -7\hat{j} \text{ km/hr (vertically downwards)} \text{The relative velocity of the rain with respect to the man } (v_{rm}) \text{ can be calculated using the formula:} \vec{v}_{rm} = \vec{v}_r - \vec{v}_m \text{Substituting the known values we get:} \vec{v}_{rm} = (-7\hat{j}) - (4\hat{i}) = -4\hat{i} - 7\hat{j} \text{The magnitude of } v_{rm} \text{ can be calculated using the Pythagorean theorem:} v_{\text{relative}} = \sqrt{7^2 + 4^2} v_{\text{relative}} = \sqrt{49 + 16} = \sqrt{65} \text{ km/hr} \text{Hence, option (2) is the correct answer.}

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
}