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Current Question (ID: 11284)

Question:

$\text{All the surfaces are smooth and springs are ideal. If a block of mass } m \text{ is given the velocity } v_0 \text{ in the right direction, then the time period of the block shown in the figure will be:}$

Options:

1. $\frac{12l}{v_0}$
2. $\frac{2l}{v_0} + \frac{3\pi}{2} \sqrt{\frac{m}{k}}$
3. $\frac{4l}{v_0} + \frac{3\pi}{2} \sqrt{\frac{m}{k}}$ (Correct)
4. $\frac{\pi}{2}\sqrt{\frac{m}{k}}$

Solution:

$\text{The time period is the time taken to complete a cycle. The motion of the block can be divided into two parts: motion in the free space and simple harmonic motion (SHM) due to the springs.}\text{Initially, the block moves a distance of } 2l \text{ to the right with constant velocity } v_0 \text{ since there is no force acting on it. The time taken for this motion is } t_1 = \frac{2l}{v_0}. \text{At this point, the block comes into contact with the spring of constant } 4k. \text{The block then undergoes half of a } \text{SHM} \text{ cycle with this spring. The time period for this } \text{SHM} \text{ is } \text{T}_1 = 2\pi\sqrt{\frac{m}{4k}} = 2\pi\frac{1}{2}\sqrt{\frac{m}{k}} = \pi\sqrt{\frac{m}{k}}. \text{The time taken for half a cycle is } t_2 = \frac{\text{T}_1}{2} = \frac{\pi}{2}\sqrt{\frac{m}{k}}. \text{After completing this half cycle, the block moves back to the center and then moves to the left a distance of } 2l \text{ with velocity } v_0 \text{ (due to conservation of energy). The time taken for this motion is } t_3 = \frac{2l}{v_0}. \text{At this point, it comes into contact with the spring of constant } k. \text{The block then undergoes half of a } \text{SHM} \text{ cycle with this spring. The time period for this } \text{SHM} \text{ is } \text{T}_2 = 2\pi\sqrt{\frac{m}{k}}. \text{The time taken for half a cycle is } t_4 = \frac{\text{T}_2}{2} = \pi\sqrt{\frac{m}{k}}. \text{Finally, the block returns to its starting position, completing one full cycle. The total time period is the sum of all these durations: } \text{T} = t_1 + t_2 + t_3 + t_4 = \frac{2l}{v_0} + \frac{\pi}{2}\sqrt{\frac{m}{k}} + \frac{2l}{v_0} + \pi\sqrt{\frac{m}{k}} = \frac{4l}{v_0} + \frac{3\pi}{2}\sqrt{\frac{m}{k}}.\text{Note: The provided solution seems to have an error. It calculates the time for a } 4l \text{ distance in one go, which isn't entirely accurate. The total distance for constant velocity motion is } 2l+2l=4l. \text{ The time for } \text{SHM} \text{ with constant } 4k \text{ is half a cycle, and the time for } \text{SHM} \text{ with constant } k \text{ is also half a cycle. So, the total time period is the sum of these parts, which is } \frac{4l}{v_0} + \frac{3\pi}{2}\sqrt{\frac{m}{k}}.\text{Hence, option (3) 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
}