Import Question JSON

Current Question (ID: 10212)

Question:
$\text{A horizontal heavy uniform bar of weight } W \text{ is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to:}$
Options:
  • 1. $W$
  • 2. $\frac{W}{2}$
  • 3. $\frac{3W}{4}$
  • 4. $\frac{W}{4}$
Solution:
\text{Initially, when both men are supporting the bar, each man supports half the weight, so each feels a force of } \frac{W}{2}. \text{When one man lets go, the bar starts to rotate about the other man's hand as a pivot point.} \text{At the instant one man lets go, we need to analyze the forces and torques about the pivot point (the remaining man's hand).} \text{The bar will have both translational and rotational motion. Using the principle of virtual work or analyzing the initial acceleration:} \text{The center of mass of the bar is at its middle, which is at a distance } \frac{L}{2} \text{ from the pivot point.} \text{At the instant of release, the bar begins to rotate. The force at the pivot must provide the centripetal acceleration for the center of mass.} \text{Using torque analysis about the pivot point and considering the initial conditions, the force on the remaining man's hand becomes } \frac{W}{4}. \text{Therefore, the correct answer is option 4: } \frac{W}{4}.

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
}