GATE 2026 IN Q.51: solution and answer

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. The pyrometer responds to the radiant energy emitted by the surface, which is fixed by the measurement. For a real surface, the total emissive power is $E=\epsilon \sigma T^4$, with $T$ in kelvin.

2. The same $E$ was used for both the assumed and the true emissivity, so ${\epsilon }_1{T}_1^4={\epsilon }_2{T}_2^4$, with ${\epsilon }_1=0.8$ and ${\epsilon }_2=0.7$.

3. Convert the calculated temperature to kelvin: $T_1=1000+273.15=1273.15\text{K}$.

4. Solve for the true temperature: $T_2=T_1{\left(\frac{{\epsilon }_1}{{\epsilon }_2}\right)}^{1/4}=1273.15{\left(\frac{0.8}{0.7}\right)}^{1/4}$.

5. Evaluate the ratio: $\ln (1.142857)\approx 0.13353$, so $(1.142857{)}^{1/4}=e^{0.03338}\approx 1.0339$.

6. $T_2\approx 1273.15\times 1.0339\approx 1316.4\text{K}$.

7. Convert back to Celsius: $T_2\approx 1316.4-273.15\approx 1043{}^{\circ }\text{C}$, which is option (A). It is higher than 1000 °C because the true emissivity is lower, so the surface must be hotter to emit the same energy.