JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Waves & Sound

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 \times 1 0^{- 2} W / m^{2}

. The value of 'a' will be

Answer: 84

Solution

The power converted to visible radiation is:

P^{'} = 0.10 \times 110 = 11 W

Intensity at distance r is

I = \frac{P ^{'}}{4 π r ^{2}}

.

The change in intensity from 1 m to 5 m is:

Δ I = \frac{P ^{'}}{4 π ( 1 ) ^{2}} - \frac{P ^{'}}{4 π ( 5 ) ^{2}} = \frac{11}{4 π} (1 - \frac{1}{25}) = \frac{11}{4 π} \times \frac{24}{25}

Calculating:

Δ I = \frac{264}{4 π \times 25} = \frac{264}{100 π} \approx \frac{264}{314.16} \approx 0.84 W/m^{2} = 84 \times 1 0^{- 2} W/m^{2}

The answer is 84.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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