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Current Question (ID: 19232)

Question:

$\text{A uniform cable of mass } M \text{ and length } L \text{ is placed on a horizontal surface such that its } \left( \frac{1}{n} \right)^{\text{th}} \text{ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface, the work done should be:}$

Options:

1. $nMgL$
2. $\frac{MgL}{2n^2}$
3. $\frac{2MgL}{n^2}$
4. $\frac{4MgL}{n^2}$

Solution:

$\text{Hint: Apply the work-energy theorem.}$ $\text{Step 1: Find the mass of the hanging part.}$ $\text{A uniform cable of mass } M \text{ and length } L \text{ is placed on a horizontal surface such that its } \left( \frac{1}{n} \right)^{\text{th}} \text{ part is hanging below the edge of the surface, then;} \Rightarrow M \propto l$ $\text{The total mass of the cable is } M \text{ and its total length is } L. \text{ Therefore, the mass per unit length is given by;} \Rightarrow M_n = \frac{M}{n}$ $\text{Step 2: Find the work done.}$ $\text{The work done } W \text{ to lift the hanging part to the surface is equal to the change in gravitational potential energy of the hanging mass.}$ $\text{The change in potential energy is given by;} \Rightarrow W = mgh$ $\text{Substituting the known values we get;} \Rightarrow W = \left( \frac{M}{n} \right) g \left( \frac{L}{2n} \right)$ $\Rightarrow W = \frac{MgL}{2n^2}$ $\text{Hence, option (2) is the correct answer.}$

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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
}