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Current Question (ID: 19972)

Question:
$\text{Two identical strings, } X \text{ and } Z, \text{ made of the same material, have tensions } T_X \text{ and } T_Z \text{ respectively. If their fundamental frequencies are } 450 \text{ Hz and } 300 \text{ Hz, respectively, the ratio } \frac{T_X}{T_Z} =$
Options:
  • 1. $0.44$
  • 2. $1.5$
  • 3. $2.25$
  • 4. $1.25$
Solution:
$\text{Hint: } f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$ $\text{Step: Find the ratio of the tensions } \frac{T_X}{T_Z}. \text{ The frequency of the wave on the string is given by:}$ $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$ $\text{As the strings are identical and made of the same material i.e., } l \text{ and } \mu \text{ are the same.}$ $f \propto \sqrt{T}$ $\Rightarrow \frac{450}{300} = \sqrt{\frac{T_X}{T_Z}}$ $\Rightarrow \sqrt{\frac{T_X}{T_Z}} = \frac{3}{2}$ $\Rightarrow \frac{T_X}{T_Z} = \frac{9}{4} = 2.25$ $\text{Hence, option (3) is the correct answer.}$

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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
}