The sun radiates electromagnetic energy at the rate of 3-9- 10 26 W- Its radius is 6-96- 10 8 m- The intensity of sunlight at the solar surface will be in W-m2 -

The correct option is A 6.4× 10 ^ 7 Intensity at solar surface=Power per unit area= WA=W4π r^2=3.9× 10^264π (6.96× 10^8)^2=6.4× 10^7W/m^2 ...

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Standard XII

Physics

Dimensional Analysis

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 sunlight at the solar surface will be (in $W / m^{2}$ ) :

6.4 \times 10^{7}

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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 A $6.4 \times 10^{7}$
Intensity at solar surface=Power per unit area= $\frac{W}{A} = \frac{W}{4 π r^{2}} = \frac{3.9 \times 10^{26}}{4 π (6.96 \times 10^{8})^{2}} = 6.4 \times 10^{7} W / m^{2}$

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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 sunlight at the solar surface will be (in W/m2) :

The correct option is A 6.4×107Intensity at solar surface=Power per unit area=WA=W4πr2=3.9×10264π(6.96×108)2=6.4×107W/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 sunlight at the solar surface will be (in $W / m^{2}$ ) :

The correct option is A $6.4 \times 10^{7}$
Intensity at solar surface=Power per unit area= $\frac{W}{A} = \frac{W}{4 π r^{2}} = \frac{3.9 \times 10^{26}}{4 π (6.96 \times 10^{8})^{2}} = 6.4 \times 10^{7} W / m^{2}$