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Current Question (ID: 9831)

Question:

$\text{In the diagram shown, force } F \text{ acts on the free end of the string. If the weight } W \text{ moves up slowly by distance } h, \text{ then work done on the weight by the string holding it will be: (pulley and string are ideal)}$

Options:

1. $Fh$
2. $2Fh$
3. $\frac{Fh}{2}$
4. $4Fh$

Solution:

$\text{Hint: Net force on the block by the string is } T = 2F.$ $\text{Step 1: Draw the free-body diagram.}$ $\text{From the pulley system shown, we can see that the string passes over a movable pulley attached to the weight. The string has tension } T \text{ throughout, and force } F \text{ is applied at the free end.}$ $\text{For the movable pulley system, two segments of the string support the weight, so the upward force on the weight is } 2T.$ $\text{Step 2: Find the relationship between applied force and tension.}$ $\text{Since the weight moves up slowly (quasi-static process), the system is in equilibrium:}$ $2T = W$ $\text{For the free end where force } F \text{ is applied: } T = F$ $\text{Therefore: } T = F \text{ and the net upward force on the weight is } 2F.$ $\text{Step 3: Find the work done by the string on the weight.}$ $\text{The string exerts an upward force of } 2F \text{ on the weight.}$ $\text{The weight moves upward by distance } h.$ $\text{Work done by the string on the weight:}$ $W = \text{Force} \times \text{displacement} = 2F \times h = 2Fh$

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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
}