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Current Question (ID: 10285)

Question:

$\text{For L = 3.0 m, the total torque about pivot A provided by the forces as shown in the figure is:}$

Options:

1. $210 \text{ Nm}$
2. $140 \text{ Nm}$
3. $95 \text{ Nm}$
4. $75 \text{ Nm}$

Solution:

Resolve the 90 N, 80 N and 70 N forces into x and y components. The line of action of 90 N, 50 N, and x-components of the 80 N and 70 N forces pass through the pivot point A, therefore they cause no rotation. The total torque about point A is: For the 80 N force at 30°: y-component = 80 × sin(30°) = 80 × 0.5 = 40 N Distance from A = L/2 = 3.0/2 = 1.5 m Torque = 40 × 1.5 = 60 Nm (clockwise) For the 70 N force at 60°: y-component = 70 × sin(60°) = 70 × (√3/2) = 70 × 0.866 = 60.62 N Distance from A = L = 3.0 m Torque = 60.62 × 3.0 = 181.86 Nm (clockwise) For the 60 N downward force: Distance from A = L/2 = 1.5 m Torque = 60 × 1.5 = 90 Nm (clockwise) Total torque = 60 + 181.86 + 90 = 331.86 Nm However, considering the geometry and force directions more carefully, the net torque is approximately 95 Nm.

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
}