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Current Question (ID: 9788)

Question:

$\text{Two plates of the same mass are attached rigidly to the two ends of a spring. One of the plates rests on a horizontal surface, and the other results in compression } X \text{ of the spring when it is in steady-state. If an external force is applied to the upper plate to just lift off the lowest plate, what further compression in the spring is required?}$

Options:

1. $0.5X$
2. $3X$
3. $2X$
4. $X$

Solution:

$\text{Hint: For the lifting, the normal reaction on the lower part is zero.}$ $\text{Step 1: Find the initial compression in the spring}$ $\text{In equilibrium: } Mg = kx_0$ $x_0 = \frac{mg}{k} \text{ (Compression state)}$ $\text{Here, } x_0 = X$ $\text{Step 2: Identify condition in which the lower plate can be lifted}$ $\text{For lifting: } kx' = mg$ $x' = \frac{mg}{k} \text{ (extension state)}$ $\text{Step 3: Find the extra compression in the spring to lift the lower plate}$ $\text{The spring must go from compression state } X \text{ to extension state } X \text{ relative to natural length.}$ $\text{Total displacement needed = } X + X = 2X$ $\text{So, further required compression is } 2X$

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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
}