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Current Question (ID: 19973)

Question:
$\text{A uniform thin rope of length } 12 \text{ m and mass } 6 \text{ kg hangs vertically from a rigid support and a block of mass } 2 \text{ kg is attached to its free end.}$ $\text{A transverse short wave-train of wavelength } 6 \text{ cm is produced at the lower end of the rope.}$ $\text{What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope?}$
Options:
  • 1. $12$
  • 2. $9$
  • 3. $3$
  • 4. $6$
Solution:
$\text{We know that the wavelength of the wave is directly proportional to the velocity of the wave i.e., } \lambda = \frac{v}{f} = \frac{\sqrt{\frac{T}{\mu}}}{f} \quad \left[ v = \sqrt{\frac{T}{\mu}} \right]$ $\Rightarrow \lambda = \sqrt{T}$ $\Rightarrow \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$ $\Rightarrow \lambda_2 = \lambda_1 \sqrt{\frac{T_2}{T_1}}$ $\Rightarrow \lambda_2 = (6 \text{ cm}) \times \sqrt{\frac{8g}{2g}} \quad \left[ T_2 = 8g (6g + 2g), T_1 = 2g \right]$ $\Rightarrow \lambda_2 = 6 \times 2 = 12 \text{ cm}$ $\text{Hence, option (1) is the correct answer.}$

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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
}