Import Question JSON

Current Question (ID: 19342)

Question:

$\text{A circular disc of mass } M \text{ and radius } R \text{ is rotating about its axis with angular speed } \omega_1. \text{ If another stationary disc having radius } \frac{R}{2} \text{ and same mass } M \text{ is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed } \omega_2. \text{ The energy lost in the process is } p\% \text{ of the initial energy. Value of } p \text{ is:}$ $1. \ 10$ $2. \ 20$ $3. \ 30$ $4. \ 40$

Options:

1. $10$
2. $20$
3. $30$
4. $40$

Solution:

$\text{Apply the conservation of angular momentum.}$ $\text{Let, the moment of inertia of bigger disc is } I = \frac{MR^2}{2}$ $\Rightarrow \text{ MOI of small disc } I_2 = \text{M} \left( \frac{R}{2} \right)^2 = \frac{I}{4}$ $\text{by angular momentum conservation}$ $I \omega_1 + \frac{1}{4} \left( 0 \right) = I \omega_2 + \frac{1}{4} \omega_2 \Rightarrow \omega_2 = \frac{4 \omega_1}{5}$ $\text{Initial kinetic energy } K_1 = \frac{1}{2} I \omega_1^2$ $\text{final kinetic energy } K_2 = \frac{1}{2} \left( I + \frac{I}{4} \right) \left( \frac{4 \omega_1}{5} \right)^2 = \frac{1}{2} \omega_1^2 \left( \frac{4}{5} \right)$ $P\% = \frac{K_1 - K_2}{K_1} \times 100\% = \frac{1 - \frac{4}{5}}{1} \times 100 = 20\%$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}