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/m². The value of 'a' will be _________. Numerical answer type. Work it out, then open the answer.

Answer: 84. Visible power P = 10% of 110 W = 11 W; I = (P)/(4π r²) I₁ = (11)/(4π (1)²) ≈ 0.875 W/m², I₅ = (11)/(4π (5)²) ≈ 0.035 W/m² Δ I = 0.875 - 0.035 = 0.84 W/m² = 84 × 10⁻² W/m² So a = 84.

Nearly 10% of the power of a 110 W light bulb is converted to visible… | JEE Main 2022 (29 Jul, Shift 2)

Q18JEE Main 2022 · 29 JUL · SHIFT 2Medium

Energy Density & Intensity

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 84

Visible power $P = 10\%$ of $110\ \text{W} = 11\ \text{W}$ ; $I = \dfrac{P}{4\pi r^2}$
$I_1 = \dfrac{11}{4\pi (1)^2} \approx 0.875\ \text{W/m}^2$ , $I_5 = \dfrac{11}{4\pi (5)^2} \approx 0.035\ \text{W/m}^2$
$\Delta I = 0.875 - 0.035 = 0.84\ \text{W/m}^2 = 84 \times 10^{-2}\ \text{W/m}^2$
So $a = 84$ .

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