Import Question JSON

Current Question (ID: 11290)

Question:

$\text{Two springs, of force constants } \text{k}_1 \text{ and } \text{k}_2 \text{ are connected to a mass } \text{m} \text{ as shown in the figure. The frequency of oscillation of the mass is } \text{f}. \text{If both } \text{k}_1 \text{ and } \text{k}_2 \text{ are made four times their original values, the frequency of oscillation will become:}$

Options:

1. $2\text{f}$ (Correct)
2. $\text{f}/2$
3. $\text{f}/4$
4. $4\text{f}$

Solution:

$\text{The springs are connected in parallel. The effective spring constant } \text{k}_{\text{eq}} \text{ for a parallel combination is } \text{k}_{\text{eq}}=\text{k}_1+\text{k}_2. \text{The frequency of oscillation of the mass is given by } \text{f}=\frac{1}{2\pi}\sqrt{\frac{\text{k}_{\text{eq}}}{\text{m}}}=\frac{1}{2\pi}\sqrt{\frac{\text{k}_1+\text{k}_2}{\text{m}}}. \text{Let the original frequency be } \text{f}_1 \text{ and the new frequency be } \text{f}_2. \text{The original values are } \text{k}_1 \text{ and } \text{k}_2. \text{The new values are } \text{k}_1'=\text{4k}_1 \text{ and } \text{k}_2'=\text{4k}_2. \text{Therefore, the new effective spring constant is } \text{k}_{\text{eq}}'=\text{k}_1'+\text{k}_2'=\text{4k}_1+\text{4k}_2=\text{4}(\text{k}_1+\text{k}_2)=\text{4k}_{\text{eq}}. \text{The new frequency is } \text{f}_2=\frac{1}{2\pi}\sqrt{\frac{\text{k}_{\text{eq}}'}{\text{m}}}=\frac{1}{2\pi}\sqrt{\frac{\text{4k}_{\text{eq}}}{\text{m}}}=2\left(\frac{1}{2\pi}\sqrt{\frac{\text{k}_{\text{eq}}}{\text{m}}}\right)=2\text{f}_1. \text{So, the new frequency of oscillation will be } 2\text{f}.$

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
}