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Current Question (ID: 11269)

Question:
$\text{A simple pendulum of mass } m \text{ swings about point } B \text{ between extreme positions } A \text{ and } C\text{. Net force acting on the bob at these three points is correctly shown by:}$
Options:
  • 1. $\text{Option 1: Forces at A and C point toward B, force at B points downward}$
  • 2. $\text{Option 2: Forces at A and C point toward B, force at B points upward}$ (Correct)
  • 3. $\text{Option 3: Forces at A and C point toward B, force at B points upward}$
  • 4. $\text{Option 4: Forces at A and C point toward B, force at B points downward}$
Solution:
$\text{Hint: Net force on a body in SHM is towards the mean position.}$ $\text{Step 1: Draw the diagram.}$ $\text{The diagram shows a pendulum with positions A, B, and C, where B is the equilibrium position and A and C are the extreme positions.}$ $\text{Step 2: Analyze the forces at each position.}$ $\text{The velocities at extreme positions A and C are zero. So there is only tangential force and no centripetal force. At B, it has centripetal force due to velocity at the mean position.}$ $\text{At positions A and C (extreme positions):}$ $\text{- Velocity } v = 0$ $\text{- Only tangential component of weight acts: } mg\cos\theta$ $\text{- This tangential force points toward the equilibrium position B}$ $\text{- No centripetal force since velocity is zero}$ $\text{At position B (equilibrium/mean position):}$ $\text{- Velocity is maximum}$ $\text{- Net force has two components:}$ $\text{1. Tangential component = 0 (since it's at equilibrium)}$ $\text{2. Centripetal component points toward the center (upward along the string)}$ $\text{Therefore, the correct answer shows forces at A and C pointing toward B (restoring force), and the force at B pointing upward (centripetal force).}$

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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
}