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PhysicsEasyNumerical2022 · 29 Jul Shift 2

Q24.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× 10-2 W m-2. The value of '𝑎' will be

What This Question Tests

This question assesses the application of the inverse square law for intensity of radiation and calculating the difference in intensity at two different distances from a point source.

Concepts Tested

PowerIntensityInverse square law

Formulas Used

I = P / (4πr^2)

ΔI = I1 - I2

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