Import Question JSON

Current Question (ID: 9695)

Question:

$\text{Rain is falling vertically with a speed of 30 m/s. A woman rides a bicycle with a speed of 10 m/s in the north to south direction. What is the direction in which she should hold her umbrella?}$ $\text{[Given: } \tan 16° = 0.29, \text{ \& } \tan 18° = 0.33\text{]}$

Options:

1. $16° \text{ with the vertical, towards north}$
2. $18° \text{ with the vertical, towards north}$
3. $16° \text{ with the vertical, towards south}$
4. $18° \text{ with the vertical, towards south}$

Solution:

$\text{Hint: In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.}$ $\text{Step 1: Draw the diagram.}$ $\text{Setting up coordinates: North-South direction as x-axis, vertical direction as y-axis.}$ $\text{The velocity vectors are:}$ $\vec{V}_R = \text{velocity of rain} = 30 \text{ m/s downward}$ $\vec{V}_G = \text{velocity of woman} = 10 \text{ m/s south}$ $\text{The relative velocity of rain with respect to woman is:}$ $\vec{V}_{R,G} = \vec{V}_R - \vec{V}_G$ $\text{Step 2: Calculate the angle.}$ $\text{From the vector diagram, the relative velocity has components:}$ $\text{Horizontal component} = V_G = 10 \text{ m/s (northward)}$ $\text{Vertical component} = V_R = 30 \text{ m/s (downward)}$ $\tan\theta = \frac{V_G}{V_R} = \frac{10}{30} = \frac{1}{3} = 0.33$ $\text{Given that } \tan 18° = 0.33$ $\text{Therefore, } \theta = 18°$ $\text{Since the woman is moving south, the relative rain velocity appears to come from the north.}$ $\text{Hence, she should hold the umbrella at } 18° \text{ with the vertical, towards north.}$

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
}