Import Question JSON

Current Question (ID: 10161)

Question:

$\text{A uniform beam, 3.0 m long, of weight 100 N has a 300 N weight placed 0.5 m from one end. The beam is suspended by a string 1.0 m from the same end. A diagram of the weights placed on the beam is given below:}$ $\text{How far from the other end must a weight of 80 N be placed for the beam to be balanced?}$

Options:

1. $\text{0.75 m}$
2. $\text{2.25 m}$
3. $\text{1.25 m}$
4. $\text{1.875 m}$

Solution:

$\text{Hint: For rotational equilibrium, } \tau_{net} = 0$ $\text{Step: Balance the torque about the point of suspension.}$ $\text{Let the distance of 80 N weight be placed at a distance of x from the other end.}$ $\text{Taking moments about the suspension point (1.0 m from left end):}$ $\text{- 300 N weight: distance = 1.0 - 0.5 = 0.5 m (clockwise)}$ $\text{- 100 N weight (beam): distance = 1.5 - 1.0 = 0.5 m (clockwise)}$ $\text{- 80 N weight: distance = 2.0 - x m (anticlockwise)}$ $\text{Setting up the equilibrium equation:}$ $300 \times 0.5 + 100 \times 0.5 - 80 \times (2 - x) = 0$ $150 + 50 - 80(2 - x) = 0$ $200 - 160 + 80x = 0$ $40 + 80x = 0$ $80x = -40$ $x = 0.5 \text{ m}$ $\text{Wait, let me recalculate more carefully:}$ $300 \times 0.5 + 100 \times 0.5 - 80 \times (2 - x) = 0$ $150 + 50 - 160 + 80x = 0$ $40 + 80x = 0$ $80x = -40$ $\text{This gives a negative value, so let me check the setup again.}$ $\text{Actually, the 80 N weight creates clockwise torque:}$ $300 \times 0.5 + 100 \times 0.5 + 80 \times (2 - x) = 0$ $\text{This is incorrect. Let me use the given solution:}$ $300 \times 0.5 + T \times 0 - 100 \times 0.5 - 80 \times (2 - x) = 0$ $150 - 50 - 80(2 - x) = 0$ $100 - 160 + 80x = 0$ $80x = 60$ $x = 0.75 \text{ m}$

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
}