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Current Question (ID: 19259)

Question:

$\text{As per the given figure, two blocks each of mass } 250 \text{ g are connected to a spring of spring constant } 2 \text{ Nm}^{-1}. \text{ If both are given velocity } v \text{ in opposite directions, then the maximum elongation of the spring is:}$

Options:

1. $\frac{v}{2\sqrt{2}}$
2. $\frac{v}{2}$
3. $\frac{v}{4}$
4. $\frac{v}{\sqrt{2}}$

Solution:

$\text{Hint: Use the conservation of the energy.}$ $\text{Step: Find the maximum elongation of the spring.}$ $\text{At maximum elongation, the velocity of both blocks will be zero.}$ $\text{By applying the conservation of energy we get:}$ $\Rightarrow \frac{1}{2} m (v)^2 \times 2 + 0 = 0 + 0 + \frac{1}{2} k x_{\text{max}}^2$ $\Rightarrow x_{\text{max}} = \sqrt{\frac{2mv^2}{k}}$ $\Rightarrow x_{\text{max}} = v \sqrt{\frac{2 \times 250}{1000 \times 2}}$ $\Rightarrow x_{\text{max}} = \frac{v}{2}$ $\text{Hence, option (2) is the correct answer.}$

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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
}