The temperature distribution over the filament is uniform.
The resistance over small sections of the filament decreases with time.
The filament emits more light at higher band of frequencies before it breaks up.
The filament consumes less electrical power towards the end of the life of the bulb.
Towards the end of the life, the filament will become thinner. Resistance will increase because resistance is inversely proportional to the cross-sectional area of the filament (R = ρL/A, where R is resistance, ρ is resistivity, L is length, and A is cross-sectional area). Since the voltage is constant (V = IR, where V is voltage, I is current, and R is resistance), an increase in resistance leads to a decrease in current. Power consumed is given by P = IV = V²/R. Since the voltage is constant and the resistance increases, the power consumed (P) will decrease. The filament will emit less light because less power is being consumed. The temperature distribution will be non-uniform because thinner sections will have higher resistance and thus higher power dissipation per unit length, leading to higher temperatures in those regions. At the position where the temperature is maximum, the filament will break due to evaporation. The black body radiation curve will shift towards lower intensity as the power decreases, but the peak will still be at higher frequencies. Therefore, statements C and D are true. Answer is C and D.