Import Question JSON

Current Question (ID: 19541)

Question:

$\text{The area of cross-section of the rope used to lift a load by a crane is } 2.5 \times 10^{-4} \text{ m}^2. \text{ The maximum lifting capacity of the crane is } 10 \text{ metric tons.}$ $\text{To increase the lifting capacity of the crane to } 25 \text{ metric tons, the required area of the cross-section of the rope should be: (Take } g = 10 \text{ ms}^{-2})$

Options:

1. $6.25 \times 10^{-4} \text{ m}^2$
2. $10 \times 10^{-4} \text{ m}^2$
3. $1 \times 10^{-4} \text{ m}^2$
4. $1.67 \times 10^{-4} \text{ m}^2$

Solution:

$\text{Hint: Breaking stress } = \frac{F}{A}$ $\text{Step: Find the required area of the cross-section of the rope.}$ $\text{Let } A \text{ be the required cross-section of the rope.}$ $\text{Since breaking stress is the property of the material so it will remain the same in both cases.}$ $\Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2}$ $\Rightarrow \frac{10}{2.5 \times 10^{-4}} = \frac{25}{A}$ $\Rightarrow A = 625 \times 10^{-6}$ $\Rightarrow 6.25 \times 10^{-4} \text{ m}^2$ $\text{Thus, the required area of cross-section of the rope is } 6.25 \times 10^{-4} \text{ m}^2.$ $\text{Hence, option (1) is the correct answer.}$

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
}