Start free

Physics · Heat Transfer

JEE Advanced 2020 — Paper 1 — Question 7

The filament of a light bulb has surface area 64 mm264 \mathrm{~mm}^{2}. The filament can be considered as a black body at

temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed

from a distance of 100 m . Assume the pupil of the eyes of the observer to be circular with radius 3 mm . Then

(Take Stefan-Boltzmann constant =5.67×10−8Wm−2 K−4=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}, Wien's displacement constant

=2.90×10−3 m−K=2.90 \times 10^{-3} \mathrm{~m}-\mathrm{K}, Planck's constant =6.63×10−34Js=6.63 \times 10^{-34} \mathrm{Js}, speed of light in vacuum

=3.00×108 ms−1=3.00 \times 108 \mathrm{~ms}^{-1} )

  1. Option A:

    power radiated by the filament is in the range 642 W to 645 W

  2. Option B:

    radiated power entering into one eye of the observer is in the range 3.15×10−8 W3.15 \times 10^{-8} \mathrm{~W} to 3.25×10−8 W3.25 \times 10^{-8} \mathrm{~W}

    Correct
  3. Option C:

    the wavelength corresponding to the maximum intensity of light is 1160 nm

    Correct
  4. Option D:

    taking the average wavelength of emitted radiation to be 1740 nm , the total number of photons entering per second into one eye of the observer is in the range 2.75×10112.75 \times 10^{11} to 2.85×10112.85 \times 10^{11}

    Correct

Answer: B, C, D

Step-by-step solution

P=AσT4\mathrm{P}=\mathrm{A} \sigma \mathrm{T}^{4}

=64×10−6×5.67×10−8×(2500)4≈140 W=64 \times 10^{-6} \times 5.67 \times 10^{-8} \times(2500)^{4} \approx 140 \mathrm{~W}

Power entering in eye

=1404π(100)2π(3×10−3)2=140×9×10−104=3.15×10−8 W=\frac{140}{4 \pi(100)^{2}} \pi\left(3 \times 10^{-3}\right)^{2}=\frac{140 \times 9 \times 10^{-10}}{4}=3.15 \times 10^{-8} \mathrm{~W}

λT=b⇒λ=2.93×10−62500⇒λ=1160\lambda \mathrm{T}=\mathrm{b} \Rightarrow \lambda=\frac{2.93 \times 10^{-6}}{2500} \Rightarrow \lambda=1160

n124201740=3.15×10−81.6×10−19=2.75×1011\mathrm{n} \frac{12420}{1740}=\frac{3.15 \times 10^{-8}}{1.6 \times 10^{-19}}=2.75 \times 10^{11}

Solution figure

Answer key and solution verified before publishing.

Practise Heat Transfer

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Physics
Chapter
Heat Transfer
Topic
Advanced Problems on Conduction

More Heat Transfer questions from this paper

The filament of a light bulb has surface area 64 mm 2 . The filament… | JEE Advanced 2020 PYQ with Solution · DhiX AI