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Current Question (ID: 11293)

Question:

$\text{A mass of } 30 \text{ g is attached with two springs having spring constant } 100 \text{ N/m and } 200 \text{ N/m and other ends of springs are attached to rigid walls as shown in the given figure. The angular frequency of oscillation will be}$

Options:

1. $\frac{100}{2\pi} \text{ rad/s}$
2. $\frac{100}{\pi} \text{ rad/s}$
3. $100 \text{ rad/s}$ (Correct)
4. $200\pi \text{ rad/s}$

Solution:

$\text{Hint: Find equivalent spring constant. Step 1: Check if springs are connected in series or parallel. In this arrangement, the springs are in parallel, so the equivalent spring constant is the sum of the individual constants. } k_{eq} = k_1 + k_2 = 100 + 200 = 300 \text{ N/m. Step 2: Find angular frequency. The formula for angular frequency is } \omega = \sqrt{\frac{k_{eq}}{m}}. \text{ Convert mass to kilograms: } 30 \text{ g} = 30 \times 10^{-3} \text{ kg. } \omega = \sqrt{\frac{300}{30 \times 10^{-3}}} = \sqrt{10^4} = 100 \text{ rad/s.}$

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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
}