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Current Question (ID: 19898)

Question:

$\text{A spring whose unstretched length is } l \text{ has a force constant } k. \text{ The spring is cut into two pieces of unstretched lengths } l_1 \text{ and } l_2 \text{ where, } l_1 = nl_2 \text{ and } n \text{ is an integer. The ratio } k_1/k_2 \text{ of the corresponding force constant, } k_1 \text{ and } k_2 \text{ will be:}$

Options:

1. $\frac{1}{n^2}$
2. $\frac{1}{n}$
3. $n^2$
4. $n$

Solution:

$\text{Hint: Spring constant is inversely proportional to the length of the spring.}$ $\frac{l}{k} = \frac{l_1}{k_1} + \frac{l_2}{k_2}$ $\text{Given } l_1 = nl_2$ $k_1 = \frac{l/l_1}{l} \times k$ $\frac{k_1}{k_2} = \frac{l_2}{l_1} = \frac{k_1}{k_2} = \frac{1}{n}$

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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
}