Import Question JSON

Current Question (ID: 9835)

Question:
$\text{A cord is used to vertically lower a block of mass m by a distance d at a constant downward acceleration of } \frac{g}{4}. \text{ The work done by the chord on the block will be:}$
Options:
  • 1. $\frac{3}{4}mgd$
  • 2. $-\frac{3}{4}mgd$
  • 3. $\frac{1}{4}mgd$
  • 4. $-\frac{1}{4}mgd$
Solution:
$\text{Let force applied by the rope } = F$ $\text{Using Newton's second law for the block moving downward with acceleration } \frac{g}{4}:$ $F - mg = -ma = -m \cdot \frac{g}{4}$ $\text{(Note: Force equation is } F - mg = -ma \text{ because acceleration is downward but rope force is upward)}$ $F = mg - \frac{mg}{4} = \frac{4mg - mg}{4} = \frac{3mg}{4}$ $\text{Work done by the rope on the block:}$ $W = F \cdot d \cdot \cos(\theta)$ $\text{Since the rope force is upward and displacement is downward, } \theta = 180°$ $\cos(180°) = -1$ $\text{Therefore:}$ $W = F \cdot d \cdot (-1) = -\frac{3mg}{4} \cdot d = -\frac{3mgd}{4}$

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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
}