Import Question JSON

Current Question (ID: 9924)

Question:

$\text{A car of mass 100 kg and traveling at 20 m/s collides with a truck weighing 1 tonne traveling at 9 km/h in the same direction. The car bounces back at a speed of 5 m/s. The speed of the truck after the impact will be:}$

Options:

1. $11.5 \text{ m/s}$
2. $5 \text{ m/s}$
3. $18 \text{ m/s}$
4. $12 \text{ m/s}$

Solution:

$\text{Let's denote:}$ $\text{Mass of car } (m_c) = 100 \text{ kg}$ $\text{Initial velocity of car } (u_c) = 20 \text{ m/s}$ $\text{Mass of truck } (m_t) = 1 \text{ tonne} = 1000 \text{ kg}$ $\text{Initial velocity of truck } (u_t) = 9 \text{ km/h}$ $\text{First, convert the truck's speed to m/s:}$ $u_t = 9 \text{ km/h} = 9 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 2.5 \text{ m/s}$ $\text{After the collision:}$ $\text{Final velocity of car } (v_c) = -5 \text{ m/s (negative because it bounces back)}$ $\text{Final velocity of truck } (v_t) = ?$ $\text{Using the principle of conservation of momentum:}$ $m_c u_c + m_t u_t = m_c v_c + m_t v_t$ $(100 \text{ kg})(20 \text{ m/s}) + (1000 \text{ kg})(2.5 \text{ m/s}) = (100 \text{ kg})(-5 \text{ m/s}) + (1000 \text{ kg})v_t$ $2000 + 2500 = -500 + 1000v_t$ $4500 = -500 + 1000v_t$ $5000 = 1000v_t$ $v_t = \frac{5000}{1000} = 5 \text{ m/s}$ $\text{The final answer is } 5 \text{ m/s}\text{.}$

Import JSON File

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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
}