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Current Question (ID: 16038)

Question:

$\text{Two identical piano wires kept under the same tension } T \text{ have a fundamental frequency of } 600 \text{ Hz. The fractional increase in the tension of one of the wires which will lead to the occurrence of } 6 \text{ beats/s when both the wires oscillate together would be:}$

Options:

1. $0.02$
2. $0.03$
3. $0.04$
4. $0.01$

Solution:

$\text{Hint: The beat frequency is given by the difference of frequencies of the piano wires.}$ $\text{Step 1: Find the relation between the tension and the frequency.}$ $\text{The frequency of the open organ pipe is given by:}$ $\nu = \frac{nv}{2L} \propto v \propto \sqrt{T}$ $\text{Step 2: Find the relative change in tension.}$ $\frac{\Delta \nu}{\nu} = \frac{1}{2} \left( \frac{\Delta T}{T} \right) \Rightarrow \left( \frac{\Delta T}{T} \right) = 2 \times \frac{\Delta \nu}{\nu} = 2 \times \frac{6}{600} = 0.02$ $\text{Therefore, the relative change in tension is } 0.02.$ $\text{Hence, option (1) is the correct answer.}$

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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
}