Verified Solution • 2025–26 Session moderate

QJEEAdv16Phc2eb: JEE Advanced 2016 Physics - Class 11 JEEAdv16Phc2eb Solution: A metal is heated in a furnace where a sensor is kept above the …

A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log⁡2(PP0){\log _2}\left( {{P \over {{P_0}}}} \right) , where P0 is a constant. When the metal surface is at a temperature of 487oC, 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 2767oC?

Step-by-Step Verified Solution

log⁡2=p1p0=1{\log _2} = {{{p_1}} \over {{p_0}}} = 1

Therefore, p1p0=2{{{p_1}} \over {{p_0}}} = 2

According to Stefan's law, p ∝\propto T2

⇒p2p1=(T2T1)4=(2767+273487+273)4=44\Rightarrow {{{p_2}} \over {{p_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{{2767 + 273} \over {487 + 273}}} \right)^4} = {4^4}

p2p1=p22p0=44⇒p2p0=2×44{{{p_2}} \over {{p_1}}} = {{{p_2}} \over {2{p_0}}} = {4^4} \Rightarrow {{{p_2}} \over {{p_0}}} = 2 \times {4^4}

log⁡2p2p0=log⁡2[2×44]{\log _2}{{{p_2}} \over {{p_0}}} = {\log _2}[2 \times {4^4}]

=log⁡22+log⁡244= {\log _2}2 + {\log _2}{4^4}

=1+log⁡228=1+8=9= 1 + {\log _2}{2^8} = 1 + 8 = 9

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