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Current Question (ID: 10774)

Question:

$\text{If } \lambda_m \text{ is the wavelength, corresponding to which the radiant intensity of a black body is at its maximum and its absolute temperature is } T\text{, then which of the following graphs correctly represents the variation of } T\text{?}$

Options:

1. $\text{Linear relationship: } \lambda_m \text{ increases linearly with } T$
2. $\text{Exponential relationship: } \lambda_m \text{ increases exponentially with } T \text{ and levels off}$
3. $\text{Exponential relationship: } \lambda_m \text{ increases exponentially with } T$
4. $\text{Inverse relationship: } \lambda_m \text{ decreases as } T \text{ increases}$ (Correct)

Solution:

\text{Hint: Using Wien's displacement law } \lambda_m \propto \frac{1}{T} \text{Explanation:} \text{Wien's displacement law states that the wavelength at which the intensity of radiation from a black body is maximum is inversely proportional to its absolute temperature.} \text{Mathematically: } \lambda_m = \frac{b}{T} \text{where } b \text{ is Wien's displacement constant } (2.898 \times 10^{-3} \text{ m·K}) \text{This means that as temperature } T \text{ increases, the peak wavelength } \lambda_m \text{ decreases, following a hyperbolic curve.} \text{Graph 4 correctly shows this inverse relationship where } \lambda_m \text{ decreases as } T \text{ increases, following the characteristic hyperbolic shape of an inverse proportion.}

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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
}