Import Question JSON

Current Question (ID: 11288)

Question:

$\text{On a smooth inclined plane, a body of mass } \text{M} \text{ is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant } \text{K}, \text{ the period of oscillation of the body (assuming the springs as massless) will be:}$

Options:

1. $2\pi \left(\frac{\text{M}}{2\text{K}}\right)^{\frac{1}{2}}$ (Correct)
2. $2\pi \left(\frac{2\text{M}}{\text{K}}\right)^{\frac{1}{2}}$
3. $2\pi \left(\frac{\text{Mg}\sin\theta}{2\text{K}}\right)$
4. $2\pi \left(\frac{2\text{Mg}}{\text{K}}\right)^{\frac{1}{2}}$

Solution:

$\text{The two springs are connected in parallel. The effective spring constant } \text{K}_{\text{eq}} \text{ for springs in parallel is given by the formula } \text{K}_{\text{eq}}=\text{K}_1+\text{K}_2. \text{Since both springs have the same spring constant } \text{K}, \text{ the effective spring constant is } \text{K}_{\text{eq}}=\text{K}+\text{K}=2\text{K}. \text{The time period of oscillation } \text{T} \text{ for a mass-spring system is given by the formula } \text{T}=2\pi \sqrt{\frac{\text{M}}{\text{K}_{\text{eq}}}}. \text{Substituting the effective spring constant, we get } \text{T}=2\pi \sqrt{\frac{\text{M}}{2\text{K}}}. \text{This can also be written as } \text{T}=2\pi \left(\frac{\text{M}}{2\text{K}}\right)^{\frac{1}{2}}.$

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
}