Import Question JSON

Current Question (ID: 19194)

Question:

$\text{One end of a massless spring of spring constant } k \text{ and natural length } l_0 \text{ is fixed while the other end is connected to a small object of mass } m \text{ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity } \omega \text{ about an axis passing through a fixed end, then the elongation of the spring will be:}$

Options:

1. $\frac{k - m\omega^2 l_0}{m\omega^2}$
2. $\frac{m\omega^2 l_0}{k + m\omega^2}$
3. $\frac{m\omega^2 l_0}{k - m\omega^2}$
4. $\frac{k - m\omega^2 l_0}{k + m\omega^2}$

Solution:

$\text{The elongation of the spring is given by the balance of forces:}$ $F = ma = m\omega^2 x$ $\text{where } x \text{ is the elongation.}$ $\text{The spring force is } F_s = kx.$ $\text{At equilibrium:}$ $k(x + l_0) = m\omega^2 x$ $\Rightarrow kx + kl_0 = m\omega^2 x$ $\Rightarrow x(k - m\omega^2) = kl_0$ $\Rightarrow x = \frac{kl_0}{k - m\omega^2}$

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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
}