Import Question JSON

Current Question (ID: 9844)

Question:

$\text{A body of mass 1 kg is thrown upwards with a velocity of 20 ms}^{-1}\text{. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10) ms}^{-2}$

Options:

1. $\text{20 J}$
2. $\text{30 J}$
3. $\text{40 J}$
4. $\text{10 J}$

Solution:

$\text{Key Idea: The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.}$ $\text{Initially, body posses only kinetic energy and after attaining a height the kinetic energy is zero.}$ $\text{Therefore, loss of energy = KE - PE}$ $= \frac{1}{2}mv^2 - mgh$ $= \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10$ $= 200 - 180$ $= 20 \text{ J}$

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
}