Najbul Gazi
Last Activity: 4 Years ago
Option -(a) r^(3/2)
The heat produce by the fuse wire= i^2R= i^2* row*L/(π*r^2)
Now, the rate of loss of heat from the fuse wire by radiation= Emissivity* Area * Excess temp. Over surrounding=e*2πrL*theta
Before the fuse will be blown out,
The heat produce in fuse wire = rate of loss of heat from the fuse
So, i^2*row*L/(πr^2) =e*2πrL*theta
i^2 = [e*2πrL*theta*πr^2]/[row*L]
Clearly, i^2 proportion r^3
And i proportion to r^(3/2)
