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Current Question (ID: 11274)

Question:

$\text{The time periods for figures (a) and (b) are } T_1 \text{ and } T_2 \text{ respectively. If all surfaces shown below are smooth, then the ratio } \frac{T_1}{T_2} \text{ will be:}$

Options:

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

Solution:

$\text{Hint: For the spring-mass system, the time period is independent of the orientation of the system.}$ $\text{Step: Find the ratio of their time periods, if all surfaces shown below are smooth.}$ $\text{The time period of the spring-mass system is given by;}$ $T = 2\pi\sqrt{\frac{m}{k}}$ $\text{Here, the two setups are identical in terms of spring constants and masses. The ratio will be:}$ $\Rightarrow \frac{T_1}{T_2} = \frac{1}{1}$ $\text{Also, the time period of the spring-mass system is independent of orientation.}$ $\Rightarrow T_1 = T_2$ $\text{Hence, option (2) is the correct answer.}$

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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
}