The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is [2025]

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Q.
The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is [2025]

1 $1.5 \times 10^{- 8} Pascals$

2 0

3 $6 \times 10^{- 8} Pascals$

4 $3 \times 10^{- 8} Pascals$

Ans.
(3)

$P_{rad} = \frac{2 I}{c}$

Where $I$ = intensity at surface

c = Speed of light

$I = \frac{Power}{Area} = \frac{450}{4 π r^{2}} = \frac{450}{4 π \times 4} = \frac{450}{16 π}$

$P_{rad} = \frac{2 \times 450}{16 π \times 3 \times 10^{8}} = \frac{150}{8 π \times 10^{8}}$

$= 5.97 \times 10^{- 8} \approx 6 \times 10^{- 8} Pascals$

Solution

Q.
The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is [2025]

1 $1.5 \times 10^{- 8} Pascals$

2 0

3 $6 \times 10^{- 8} Pascals$

4 $3 \times 10^{- 8} Pascals$

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Solution

Q. The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is [2025]

1 1.5×10–8 Pascals

2 0

3 6×10–8 Pascals

4 3×10–8 Pascals

Ans. (3) Prad=2Ic Where I = intensity at surface c = Speed of light I=PowerArea=4504πr2=4504π×4=45016π Prad=2×45016π×3×108=1508π×108 =5.97×10–8≈6×10–8 Pascals

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Q.
The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is [2025]

1 $1.5 \times 10^{- 8} Pascals$

3 $6 \times 10^{- 8} Pascals$

4 $3 \times 10^{- 8} Pascals$