The sun radiates electromagnetic energy at the rate of 3-9 - 1026W- Its radius is 6-96- 108m-The intensity of sun light at the solar surface will be -in W-m2

The correct option is D 5.6 × 10^7 I= P / 4π R^ 2 = 3.9× 10 ^ 26 / 4× 3.14×( 6.96× 10 ^ 8 ) #160; ^ 2 =5.6× 10^ 7 w/ m^ 2 ...

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Standard XII

Physics

Kepler's Law

The sun radia...

Question

The sun radiates electromagnetic energy at the rate of $3.9 \times 10^{26} W$ . Its radius is $6.96 \times 10^{8} m$ .The intensity of sun light at the solar surface will be $- (i n W / m^{2})$

1.4 \times 10^{4}

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2.8 \times 10^{5}

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4.2 \times 10^{6}

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5.6 \times 10^{7}

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Solution

The correct option is D $5.6 \times 10^{7}$
$I = \frac{P}{4 π R^{2}} = \frac{3.9 \times 10^{26}}{4 \times 3.14 \times {(6.96 \times 10^{8})}^{2}} = 5.6 \times 10^{7} w / m^{2}$

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The sun radiates electromagnetic energy at the rate of 3.9×1026W. Its radius is 6.96×108m.The intensity of sun light at the solar surface will be −(inW/m2)

The correct option is D 5.6×107I=P4πR2=3.9×10264×3.14×(6.96×108)2=5.6×107w/m2

The sun radiates electromagnetic energy at the rate of $3.9 \times 10^{26} W$ . Its radius is $6.96 \times 10^{8} m$ .The intensity of sun light at the solar surface will be $- (i n W / m^{2})$

The correct option is D $5.6 \times 10^{7}$
$I = \frac{P}{4 π R^{2}} = \frac{3.9 \times 10^{26}}{4 \times 3.14 \times {(6.96 \times 10^{8})}^{2}} = 5.6 \times 10^{7} w / m^{2}$