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Current Question (ID: 20007)

Question:

$\text{A wire of length } 30 \text{ cm, stretched between rigid supports, has its } n^{\text{th}}$ $\text{and } (n+1)^{\text{th}} \text{ harmonics at } 400 \text{ Hz and } 450 \text{ Hz, respectively. If the tension in the string is } 2700 \text{ N, its linear mass density is:}$

Options:

1. $3 \text{ kg/m}$
2. $6 \text{ kg/m}$
3. $9 \text{ kg/m}$
4. $1 \text{ kg/m}$

Solution:

$\text{The frequency of the } n^{\text{th}} \text{ harmonic is given by } f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$ $\text{and for the } (n+1)^{\text{th}} \text{ harmonic, } f_{n+1} = \frac{n+1}{2L} \sqrt{\frac{T}{\mu}}$ $\text{Given } f_n = 400 \text{ Hz and } f_{n+1} = 450 \text{ Hz}$ $\text{Thus, } \frac{n+1}{n} = \frac{450}{400} = \frac{9}{8}$ $\Rightarrow n = 8 \text{ and } n+1 = 9$ $\text{Using } f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \Rightarrow 400 = \frac{8}{2 \times 0.3} \sqrt{\frac{2700}{\mu}}$ $\Rightarrow \mu = 3 \text{ kg/m}$

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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
}