Question

Two electric bulbs have filaments of lengths $L$ and $2L$, diameters $2d$ and $d$ and emissivities $3e$ and $4e$. If their temperatures are in the ratio $2:3$, their powers will be in the ratio of

• A. 8 : 27
• B. 4 : 27
• C. 8 : 3
• D. 4 : 9

Answer 1

Answer: (B)

4 : 27

We know that emission from the surface is given by following relation:
$E=\epsilon \sigma A{T}^4$

$\frac{E_1}{E_2}=\frac{{\epsilon }_1{A}_1{T}_1^4}{{\epsilon }_2{A}_2{T}_2^4}$

$\frac{E_1}{E_2}=\frac{3\epsilon \times 2d\times L\times 2^4}{4\epsilon \times d\times 2L\times 3^4}=\frac{4}{27}$