Import Question JSON

Current Question (ID: 19251)

Question:

$\text{A ball with a mass of } 100 \text{ g is dropped from a height of } h = 10 \text{ cm}$ $\text{onto a platform fixed at the top of a vertical spring (as shown in the figure).}$ $\text{The ball remains on the platform, and the platform is}$ $\text{depressed by a distance of } \frac{h}{2}. \text{ The spring constant is: (use } g = 10 \text{ ms}^{-2})$

Options:

1. $100 \text{ Nm}^{-1}$
2. $110 \text{ Nm}^{-1}$
3. $120 \text{ Nm}^{-1}$
4. $130 \text{ Nm}^{-1}$

Solution:

$\text{At point } A, \ K.E = 0, \ P.E = mgh, \ E_{\text{spring}} = 0$ $\text{At point } B, \ K.E = 0, \ P.E = -mg\frac{h}{2}, \ E_{\text{spring}} = \frac{1}{2}kx^2$ $\text{By applying energy conservation at point } A \text{ and } B \text{ we get;} \ mgh = \frac{1}{2}kx^2 - mg\frac{h}{2}$ $\Rightarrow mg \left( h + \frac{h}{2} \right) = \frac{1}{2}kx^2$ $\Rightarrow mg \left( \frac{3h}{2} \right) = \frac{1}{2}kx^2$ $\Rightarrow 0.100 \times 10 \times \frac{3}{2} (0.10) = \frac{1}{2}k(0.05 \times 0.05) \quad \left[ x = \frac{h}{2} \right]$ $\Rightarrow k = \frac{\frac{3 \times 0.10}{0.05 \times 0.05}}{\frac{3 \times 1000}{25}}$ $\Rightarrow k = 120 \text{ N/m}$ $\text{Hence, option (3) is the correct answer.}$

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
}