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Current Question (ID: 11276)

Question:

$\text{The period of oscillation of a mass } M \text{ suspended from a spring of negligible mass is } T. \text{ If along with it, another mass } M \text{ is also suspended, the period of oscillation will now be:}$

Options:

1. $T$
2. $\frac{T}{\sqrt{2}}$
3. $2T$
4. $\sqrt{2}T$ (Correct)

Solution:

$\text{Hint: The time period of oscillation, } T = 2\pi\sqrt{\frac{m}{k}}$ $\text{Step 1: Find the initial time period.}$ $\text{The initial time period, } T = 2\pi\sqrt{\frac{M}{k}}$ $\text{Step 2: Find the final time period.}$ $\text{The final time period,}$ $T' = 2\pi\sqrt{\frac{M+M}{k}} = 2\pi\sqrt{\frac{2M}{k}}$ $\text{Step 3: Find the ratio of the time periods.}$ $\text{Now, } \frac{T'}{T} = \frac{\sqrt{2}}{1}$ $\Rightarrow T' = \sqrt{2}T$

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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
}