Explanation

(b) An electric fuse of length $l$, radius $r$ when used in series of the circuit can withstand only if the rate of heat produced due to current in it is equal to the rate of heat lost due to radiation. If $H$ is the rate of lost per unit area of the fuse wire, then $H \times 2\pi rl = I^{2}R = I^{2}\rho l/\pi r^{2}$
or $H = \frac{I^{2}\rho}{2\pi^{2}r^{3}}\ ie,\ H$ is independent of $l$.