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Current Question (ID: 10775)

Question:
$\text{Three stars A, B, and C have surface temperatures } T_A, T_B \text{ and } T_C \text{ respectively. Star A appears bluish, star B appears reddish and star C yellowish. Hence:}$
Options:
  • 1. $T_A > T_B > T_C$
  • 2. $T_B > T_C > T_A$
  • 3. $T_C > T_B > T_A$
  • 4. $T_A > T_C > T_B$ (Correct)
Solution:
\text{Hint: The colours of the stars depend on their temperature.} \text{Step 1: Apply Wien's law to find the relation between the temperatures.} \text{Wien's law is: } \lambda_m T = c \text{Therefore: } T = \frac{c}{\lambda_m} \text{ or } T \sim \frac{1}{\lambda_m} \text{The visible light spectrum shows: V I B G Y O R with wavelength increasing from left to right} \text{Therefore: wavelength increases } \lambda \text{ increases} \text{And: temperature decreases } T \text{ decreases} \text{Step 2: Find the relation between the temperatures.} \text{According to Wien's law and the color spectrum:} \text{Blue light has shorter wavelength than yellow, and yellow has shorter wavelength than red} \text{Since } T \sim \frac{1}{\lambda_m}\text{, shorter wavelength means higher temperature} \text{Therefore: Bluish star A has highest temperature, yellowish star C has medium temperature, and reddish star B has lowest temperature} T_A > T_C > T_B

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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
}