This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Remote Sensing – Characteristics of Solar Radiation”.

1. Diameter of sun can be given as ____________

a) 1.39 * 10^{7} km

b) 1.9 * 10^{6} km

c) 1.39 * 10^{6} km

d) 1.39 * 10^{16} km

View Answer

Explanation: Sun is the most prominent star among all the stars available. The radiation emitted by sun can used in case of passive remote sensing. The diameter of the sun can be estimated as 1.39 * 10

^{6}km.

2. The energy radiated from sun in visible region will be around ____________

a) 43%

b) 45%

c) 47%

d) 50%

View Answer

Explanation: The radiation emitted from sun will be in a huge manner. Emitted radiation can be visualized up to some extent and remaining is not visible to naked eye. It can be estimated that 43% can be visualized and remaining 48% is transferred to IR region.

3. The value of solar constant can be given as __________

a) 1637 W / m^{2}

b) 1367 W / m^{2}

c) 136 W / m^{2}

d) 3167 W / m^{2}

View Answer

Explanation: The energy received from the sun can be distributed all over the surface of earth and by doing this an average incident flux density can be established. It is determined as solar constant, having a value of 1367 W / m

^{2}.

4. Temperature on the sun is around ______________

a) 575 – 600 K

b) 7550 – 8000 K

c) 5570 – 6000 K

d) 5750 – 6000 K

View Answer

Explanation: Sun, being the most vulnerable among all the stars, is set to emit a lot of radiation. In order to emit such radiation, it must possess a large amount of temperature. The temperature range would be around 5750 – 6000 K.

5. Around how much percentage, the incident radiant flux can be absorbed by the materials present on earth?

a) 48%

b) 37%

c) 42%

d) 50%

View Answer

Explanation: The radiation which is being spread around all over the world is set to possess an incident flux density of 1367 W / m

^{2}. This is divided into certain classes in which 48% of this is absorbed by the earth’s surface materials.

6. Determine the spectral existence of a black body if the absolute temperature can be given as 300K.

a) 1968*10^{-11} W/sq. m

b) 1689*10^{-11} W/sq. m

c) 6298*10^{-11} W/sq. m

d) 1698*10^{-11} W/sq. m

View Answer

Explanation: By using Stefan – Boltzmann law, the spectral existence of a black body can be determined. The formula can be given as M = σ*T

^{4}. On substitution, we get

M = 5.66*10

^{-11}* 300

M = 1698*10

^{-11}W/sq. m.

7. Find the value of λ, if the temperature of the body is given as 560K.

a) 4.16*10^{-14} m

b) 6.16*10^{-14} m

c) 5.16*10^{-14} m

d) 5.16*10^{-4} m

View Answer

Explanation: From Wien’s displacement law,

λ = A / T. here, A is the Wien’s constant = 2.89*10

^{-11}mK. On substitution, we get

λ = 2.89*10

^{-11}/ 560

λ = 5.16*10

^{-14}m.

8. Determine the spectral existence of a body by using plank’s law, if the wave length is given as 456m and the absolute temperature is 765K.

a) 3.6*10^{30} W / sq. m

b) 4.6*10^{30} W / sq. m

c) 6.6*10^{30} W / sq. m

d) 3.6*10^{30} W / sq. m

View Answer

Explanation: From the plank’s law, \(M = C_1 / (λ^5*e^{(\frac{c_2}{λT})-1})\). Here, C1 and C2 are the radiation constants. On substitution, we get

M = 3.74*10

^{16}/ \((765^5*e^{(1.43*\frac{10^2}{456*765})-1})\)

M = 3.6*10

^{30}W / sq. m.

9. Determine the absolute temperature of the body from Stefan – Boltzmann law if spectral existence of the body is 3.55 * 10^{-10} W /sq. m.

a) 19.58 K

b) 1.58 K

c) 15.58 K

d) 1.85 K

View Answer

Explanation: The Stefan – Boltzmann law can be given as, M = σ*T

^{4}. On substitution, we get

3.55*10

^{-10}= 5.66*10

^{-11}* T

^{4}

T = 1.58 K.

10. From Wien’s displacement law, determine the absolute temperature if wave length is given as 0.05m.

a) 7.8*10^{-3} K

b) 75.8*10^{-3} K

c) 57.25*10^{-35} K

d) 57.8*10^{-3} K

View Answer

Explanation: The formula for Wien’s displacement law can be given as

λ = A / T. On substitution, we get

0.05 = 2.89*10

^{-3}/ T

T = 57.8*10

^{-3}K.

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