Question

Two spheres ‘S₁’ and ‘S₂’ have same radii but temperatures are ‘T₁’ and ‘T₂’ respectively. Their emissive power is same and emissivity is in the ratio 1 : 4. Then the ratio ‘T₁’ to‘T₂’ is

• A. 1 : 2
• B. 2 : 1
• C. √2 : 1
• D. 1 : √2

Answer 1

The Correct Option is C

The emissive power can be represented by the Stefan-Boltzmann law: 
E = σ * ε * T⁴ 
Given that the emissive power is the same for both spheres S₁ and S₂, and the emissivity ratio is 1:4, we can write: 
E₁ = E₂ 
σ * ε₁ * T₁⁴ = σ * ε₂ * T₂⁴ 
ε₁ * T₁⁴ = ε₂ * T₂⁴ 
Since the ratio of emissivity (ε₁ : ε₂) is given as 1:4, we can substitute ε₁ = x and ε₂ = 4x, where x is a constant: 
x * T₁⁴ = 4x * T₂⁴ 
Dividing both sides by x, we get: 
T₁⁴ = 4T₂⁴ 

\(\frac {T_1^4}{T_2^4}\) = 4
Taking the fourth root of both sides, we have: 
T₁ = √2 * T₂ 
Therefore, the ratio of T₁ to T₂ is (C) √2 : 1.

Answer 2

Given:
Emissive power \(E = \sigma \epsilon T^4\)
Emissivity ratio \(\epsilon_1 : \epsilon_2 = 1 : 4\)
Since the emissive power E is the same for both spheres:
\(\sigma \epsilon_1 T_1^4 = \sigma \epsilon_2 T_2^4\)
Canceling \(\sigma\) from both sides (since it's common):
\(\epsilon_1 T_1^4 = \epsilon_2 T_2^4\)

Given \(\epsilon_1 = x\) and \(\epsilon_2 = 4x\) (where x is a constant):
\(x T_1^4 = 4x T_2^4\)

Divide both sides by x:
\(T_1^4 = 4 T_2^4\)
Taking the fourth root of both sides:
\(T_1 = \sqrt[4]{4} \cdot T_2\)
\(T_1 = \sqrt{2} \cdot T_2\)
 \(\frac{T_1}{T_2}=\) \(\frac{\sqrt{2}}{1}\)

So, the correct option is (C): \(\sqrt{2}:1\)