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Current Question (ID: 9744)

Question:
$\text{A 5 m long uniformly thick string rests on a horizontal frictionless surface. It is pulled by a horizontal force of 5 N from one end. The tension in the string at 1 m from the end where the force is applied is:}$
Options:
  • 1. $\text{zero}$
  • 2. $5\text{ N}$
  • 3. $4\text{ N}$
  • 4. $1\text{ N}$
Solution:
\text{Hint: Apply Newton's second law of motion.} \text{Step 1: Draw FBD} \text{String length = 5 m, Applied force = 5 N} \text{Find tension 1 m from force application point} \text{Step 2: Use } F = ma \text{For entire string: } 5 = Ma \text{So: } a = \frac{5}{M} \text{For 4 m segment (from point to free end):} \text{Mass of segment = } \frac{4M}{5} \text{Newton's law: } 5 - T = \frac{4M}{5} \times a \text{Step 3: Solve for T} 5 - T = \frac{4M}{5} \times \frac{5}{M} = 4 T = 5 - 4 = 1 \text{ N} \text{Therefore, the tension is 1 N.}

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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
}