Thermal Properties of Matter

JEE Main level, numerical answer, medium

119 of 126

Question 119 of 126

Nearly 1010% of the power of a 110W110W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1m1m from the bulb to a distance of 5m5m is a×10−2Wm−2a\times {10}^{-2}W{m}^{-2}. The value of 'aa' will be

Show answer and solution

Answer: 84

Power of visible radiation is P′=10100×110W{P}^{'}=\dfrac{10}{100}\times 110W =11W=11W
Change in average intensity of visible radiation is ∆Iradiation=Iradiation1−Iradiation2∆{I}_{radiation}={I}_{{radiation}_{1}}-{I}_{{radiation}_{2}}
⇒I1−I2=P′4πr12−P′4πr22\Rightarrow {I}_{1}-{I}_{2}=\dfrac{{P}^{'}}{4\pi {r}_{1}^{2}}-\dfrac{{P}^{'}}{4\pi {r}_{2}^{2}}
=114π[11−125]=\dfrac{11}{4\pi }\left[\dfrac{1}{1}-\dfrac{1}{25}\right] =114π×2425=\dfrac{11}{4\pi }\times \dfrac{24}{25}
=264π×10−2=84×10−2Wm−2=\dfrac{264}{\pi }\times {10}^{-2}=84\times {10}^{-2}W{m}^{-2}
Hence, the value of a=84a=84.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermal Properties of Matter (Radiation), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.