Import Question JSON

Current Question (ID: 19635)

Question:

$\text{At what temperature a gold ring of diameter } 6.230 \text{ cm be heated so that it can be fitted on a wooden bangle of diameter } 6.241 \text{ cm? Both diameters have been measured at room temperature } (27^\circ\text{C}). \text{ (Given: coefficient of linear thermal expansion of gold, } \alpha_L = 1.4 \times 10^{-5}\text{K}^{-1})$

Options:

1. $125.7^\circ\text{C}$
2. $91.7^\circ\text{C}$
3. $425.7^\circ\text{C}$
4. $152.7^\circ\text{C}$

Solution:

$\text{Hint: } L = L_0(1 + \alpha_L \Delta T)$ $\text{Step: Find the required temperature.}$ $\text{The length at temperature } T \text{ is given by:}$ $\Rightarrow L = L_0(1 + \alpha_L \Delta T) \quad \cdots (1)$ $\text{The change in diameter needed for the ring to fit over the bangle is given by:}$ $\Rightarrow \Delta D = D_{\text{bangle}} - D_{\text{ring}} = 6.241 \text{ cm} - 6.230 \text{ cm} = 0.011 \text{ cm}$ $\text{Using (1), we get:}$ $\Rightarrow \Delta D = D_{\text{ring}} \cdot \alpha_L \cdot \Delta T$ $\Rightarrow \Delta T = \frac{\Delta D}{D_{\text{ring}} \cdot \alpha_L}$ $\text{Substituting the known values, we get:}$ $\Rightarrow \Delta T = \frac{0.011 \text{ cm}}{6.230 \text{ cm} \times (1.4 \times 10^{-5} \text{ K}^{-1})}$ $\Rightarrow \Delta T = \frac{0.011}{8.722 \times 10^{-5}} \approx 125.7^\circ\text{C}$ $\text{Thus, the final temperature is:}$ $\Rightarrow 125.7^\circ\text{C} + 27^\circ\text{C} = 152.7^\circ\text{C}$ $\text{Hence, option (4) is the correct answer.}$

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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
}