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← JEE Main 2022 · 29 Jul · Shift 2
PhysicsInteger+4 / -1Question 52 of 90

JEE Main 2022 · 29 Jul · Shift 2, Physics question 52

Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a×10−2a\times10^{-2} W/m2^2. The value of 'a' will be _______.

Answer and worked solution

Official answer: 84

Intensities of 11 W visible: 114π(1−125)≈0.84\dfrac{11}{4\pi}\left(1-\dfrac1{25}\right)\approx0.84 W/m² =84×10−2=84\times10^{-2}.

Answer: 84