Due to a fall in mains voltage, the temperature of the filament of a bulb falls from 3000 K to 2000 K. What is the percentage fall in power consumption?
Physics · Mixed Chapter
JEE Main Practice

Question

Due to a fall in mains voltage, the temperature of the filament of a bulb falls from 3000 K to 2000 K. What is the percentage fall in power consumption?

G1 Physics · Accepted Answer

Step 1: For the expected blackbody-radiation model, radiated power P ∝ T⁴ (assuming surface area and emissivity constant). Step 2: P₂/P₁ = (T₂/T₁)⁴ = (2000/3000)⁴ = (2/3)⁴ = 16/81. Step 3: Percentage fall = (1 − 16/81) × 100 = (65/81) × 100 ≈ 80.2469%. Final answer: 80.25%. Note: The source treats electrical power consumption as radiated blackbody power; real lamp efficiency/conduction effects are ignored.