In thermal radiation for a black body (where, ε is emissivity and α is absorptivity)-A-α = 1; ε ≠ 1B-α ≠ 1, ε = 1C-α ≠ 1, ε ≠ 1D-α = 1; ε = 1-D-α = 1; ε = 1

Question

In thermal radiation for a black body (where, &epsilon; is emissivity and &alpha; is absorptivity)-A-&alpha; = 1; &epsilon; &ne; 1B-&alpha; &ne; 1, &epsilon; = 1C-&alpha; &ne; 1, &epsilon; &ne; 1D-&alpha; = 1; &epsilon; = 1-D-&alpha; = 1; &epsilon; = 1

APTITUDE CRACK · Accepted Answer

D &alpha; = 1; &epsilon; = 1 In thermal radiation for a black body, both the absorptivity (&alpha;) and emissivity (&epsilon;) are equal to 1. So, the correct statement is: d) =1;=1&alpha;=1;&epsilon;=1

Heat Transfer - Engineering

In thermal radiation for a black body (where, ε is emissivity and α is absorptivity)

For help Students Orientation
Mcqs Questions

Quick Links

Newsletter

Heat Transfer - Engineering

In thermal radiation for a black body (where, ε is emissivity and α is absorptivity)

Solution

For help Students Orientation Mcqs Questions

For help Students Orientation
Mcqs Questions