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← JEE Advanced 2016 · Paper 1
PhysicsInteger+3Question 14 of 54

JEE Advanced 2016 · Paper 1, Physics question 14

A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P)(P) by the metal. The sensor has a scale that displays log⁡2(P/P0)\log_2(P/P_0), where P0P_0 is a constant. When the metal surface is at a temperature of 487 °C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 °C?

Answer and worked solution

Official answer: 9

Temperatures. T1=487+273=760T_1=487+273=760 K and T2=2767+273=3040T_2=2767+273=3040 K =4T1=4T_1.

Stefan's law at constant emissivity.
P2P1=44=28.\frac{P_2}{P_1}=4^4=2^8.

Reading. log⁡2(P1/P0)=1\log_2(P_1/P_0)=1, so
log⁡2(P2/P0)=1+8=9.\log_2(P_2/P_0)=1+8=9.

Answer: 9