Q154 / 160 · TS EAMCET 2021 (04 Aug Shift 2) · Physics

If $10 %$ of the power of a $(100 π) W$ light bulb is converted to visible radiation, then the average intensity of visible radiation at a distance of $10 m$ is

Correct answer
A
$0.025 W m^{- 2}$
B
$0.01 W m^{- 2}$
C
$0.031 W m^{- 2}$
D
$0.05 W m^{- 2}$

Solution

The intensity is defined as power radiated per unit area here power, $P = 100 π W$ , but in the question we want to calculate only for ten percent of origin it means, $P' = 10 π W$ , so the intensity,
$I = \frac{P'}{4 π d^{2}} = \frac{10 π}{4 π {(10)}^{2}} = 0.025 W m^{- 2}$ .

Hence, the correct answer is option: 'A'.

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If 10% of the power of a 100π W light bulb is converted to visible radiation, then the average intensity of visible radiation at a distance of 10 m is

Solution

Relevant Questions

If $10 %$ of the power of a $(100 π) W$ light bulb is converted to visible radiation, then the average intensity of visible radiation at a distance of $10 m$ is