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Current Question (ID: 9839)

Question:

$\text{A force } F = (20 + 10y) \text{ acts on a particle in the } y\text{-direction where } F \text{ is in Newton and } y \text{ is in meter. Work done by this force to move the particle from } y = 0 \text{ to } y = 1 \text{ m is:}$

Options:

1. $20 \text{ J}$
2. $30 \text{ J}$
3. $5 \text{ J}$
4. $25 \text{ J}$

Solution:

$\text{Given: } F = 20 + 10y$ $\text{For work done by a variable force, we use:}$ $dW = F \cdot dy$ $dW = (20 + 10y) dy$ $\text{Integrating from } y = 0 \text{ to } y = 1:$ $W = \int_0^1 (20 + 10y) dy$ $W = [20y + 5y^2]_0^1$ $W = (20 \cdot 1 + 5 \cdot 1^2) - (20 \cdot 0 + 5 \cdot 0^2)$ $W = 20 + 5 = 25 \text{ J}$ $\text{Hence, option (4) is the correct answer.}$

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
}