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Current Question (ID: 10351)

Question:

The length of elastic string, obeying Hooke's law is $l_1$ metres when the tension is $4 \text{ N}$, and $l_2$ metres when the tension is $5 \text{ N}$. The length in metres when the tension is $0 \text{ N}$ will be:

Options:

1. $5l_1 - 4l_2$
2. $5l_2 - 4l_1$
3. $9l_1 - 8l_2$
4. $9l_2 - 8l_1$

Solution:

According to Hooke's law, the tension $F$ in an elastic string is proportional to the extension $(l - l_0)$, where $l_0$ is the original length of the string and $l$ is the stretched length. So, $F = k(l - l_0)$, where $k$ is the spring constant. Let the original length of the elastic string be $l_0$. When the tension is $4 \text{ N}$, the length is $l_1$ metres: $4 = k(l_1 - l_0) \quad \ldots (1)$ When the tension is $5 \text{ N}$, the length is $l_2$ metres: $5 = k(l_2 - l_0) \quad \ldots (2)$ Divide equation (1) by equation (2): $\frac{4}{5} = \frac{k(l_1 - l_0)}{k(l_2 - l_0)}$ $\frac{4}{5} = \frac{l_1 - l_0}{l_2 - l_0}$ Cross-multiply: $4(l_2 - l_0) = 5(l_1 - l_0)$ $4l_2 - 4l_0 = 5l_1 - 5l_0$ Rearrange the terms to solve for $l_0$: $5l_0 - 4l_0 = 5l_1 - 4l_2$ $l_0 = 5l_1 - 4l_2$ Thus, the length in metres when the tension is $0 \text{ N}$ (which is the original length) is $5l_1 - 4l_2$. Therefore, the correct answer is option (1).

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}