$\frac{80-70}{12}=\mathrm{K}\left(\frac{80+70}{2}-25\right) \ldots(1)$
$\frac{70-60}{t}=K\left(\frac{70+60}{2}-25\right) \ldots(2)$
on solving $: t=15 \mathrm{min}$
($A$) The temperature distribution over the filament is uniform
($B$) The resistance over small sections of the filament decreases with time
($C$) The filament emits more light at higher band of frequencies before it breaks up
($D$) The filament consumes less electrical power towards the end of the life of the bulb