JEE Main 2025 Apr 7 Shift 1, Physics Q24: Newton’s Law of Cooling

Setup. Convert units and compute surface area of the cylindrical filament.

$T=1727^{\circ }\text{C}+273=2000$ K

$L=0.10$ m, diameter $d=0.5$ mm $=5\times 10^{-4}$ m

Surface area (lateral): $A=\pi dL=3.14\times 5\times 10^{-4}\times 0.10=1.57\times 10^{-4}{\text{ m}}^2$

Stefan-Boltzmann law:

$P=\epsilon \sigma A{T}^4$

Rearrange for $\epsilon$:

$\epsilon =\frac{P}{\sigma A{T}^4}$

Compute $T^4$:

$T^4=(2000{)}^4=1.6\times 10^{13}{\text{ K}}^4$

Compute the denominator:

$\sigma A{T}^4=(6.0\times 10^{-8})(1.57\times 10^{-4})(1.6\times 10^{13})=(6.0\times 1.57\times 1.6)\times 10^{-8-4+13}$

$=15.072\times 10^1=150.72\text{ W}$

Emissivity:

$\epsilon =\frac{94.2}{150.72}\approx 0.625=\frac{5}{8}$

Matching with $x/8$: $x=5$.

$\therefore$ Answer = 5.