*Question*

The dashed line on the graph shows the variation with wavelength of the intensity of solar radiation before passing through the Earth’s atmosphere.

The solid line on the graph shows the variation with wavelength of the intensity of solar radiation after it has passed through the Earth’s atmosphere.

Which feature of the graph helps explain the greenhouse effect?

A. Infrared radiation is absorbed at specific wavelengths.

B. There is little absorption at infrared wavelengths.

C. There is substantial absorption at visible wavelengths.

D. There is little absorption at UV wavelengths.

**Answer/Explanation**

### Markscheme

A

**The Atmosphere And Air Mass**

The atmosphere scatters and absorbs some of the Sun’s energy that is incident on the Earth’s surface. Scattering of radiation by gaseous molecules (e.g. O_{2}, O_{3}, H_{2}O and CO_{2}), that are a lot smaller than the wavelengths of the radiation, is called *Rayleigh scattering*. Roughly half of the radiation that is scattered is lost to outer space, the remaining half is directed towards the Earth’s surface from all directions as *diffuse radiation*. Because of absorption by oxygen and ozone molecules the shortest wavelength that reaches the Earth’s surface is approximately 0.29 µm. Other gas molecules absorbed difference wavelengths as indicated in figure below.

*Question*

The average albedo of glacier ice is 0.25.

What is \(\frac{{{\text{power absorbed by glacier ice}}}}{{{\text{power reflected by glacier ice}}}}\)?

A. 0.25

B. 0.33

C. 2.5

D. 3.0

**Answer/Explanation**

### Markscheme

D

\(albedo =\alpha =\frac{\text{total reflected power}}{\text{total incident power}}\)

Here you can see total incident power = total reflected power + Total absorbed power

Incident Power = Reflected + absorbed

Now

\(albedo =\alpha =\frac{\text{total reflected power}}{\text{total incident power}}\)

\(0.25 = \frac{\text{total reflected power}}{\text{total incident power}}\)

\(0.25 =\frac{reflected}{reflected + absorbed} \) Since incident =reflected + absorbed

\(0.25 \times (reflected + absorbed) =reflected\)

\(0.25 \times (absorbed) =0.75 times reflected\)

\(\frac{absorbed}{reflected} =\frac{0.75}{0.25}=3\)

*Question*

An object can lose energy through

I. conduction

II. convection

III. radiation

What are the principal means for losing energy for a hot rock resting on the surface of the Moon?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

**Answer/Explanation**

## Markscheme

B

*Question*

X and Y are two spherical black-body radiators that emit the same total power. The absolute temperature of X is half that of Y.

What is \(\frac{{{\text{radius of X}}}}{{{\text{radius of Y}}}}\)?

A. 4

B. 8

C. 16

D. 32

**Answer/Explanation**

### Markscheme

A

Black-body radiation The power radiated by a body is governed by Stefan-Boltzman law

\(P=\varepsilon \sigma AT^4\)

Where

\(P= power\)

\(\varepsilon =emissivity = 1 \) for Black body

\(\sigma =\) Stefan-Boltzmann constant

\(A= Area = 4\pi r^2\)

T = temperature

\(\frac{P_X}{P_Y} =\frac{\varepsilon \sigma A_X T_X1^4}{\varepsilon \sigma A_Y T_Y^4}\)

\(=\frac{\varepsilon \sigma 4\pi R_X^2 T_X^4}{\varepsilon \sigma 4\pi R_Y^2T_Y^4}\)

Given \(X\) and \(Y\) to be black body and \( P_X =P_Y\) and \(T_X = \frac{T_Y}{2}\)

hence \(\varepsilon =1\)

Putting these values in above equation , we get

\(\frac{ R_X^2 T_X^4}{ R_Y^2T_Y^4} =1\)

or

\(\frac{R_X^2}{R_Y^2} =\frac{T_Y^4}{T_X^4}=2^4\)

or

\(\frac{R_X}{R_Y}=2^2=4\)

*Question*

The solar constant is the intensity of the Sun’s radiation at

A. the surface of the Earth.

B. the mean distance from the Sun of the Earth’s orbit around the Sun.

C. the surface of the Sun.

D. 10km above the surface of the Earth.

**Answer/Explanation**

## Markscheme

B

#### SOLAR CONSTANT

\(P= A\sigma T^4\)

\(\frac{P}{A} = 5.67\times 10^{-8} \times (5800)^4\)

\(=6.42\times 10^7\) Watt

Radius of \(Sun = 6.9 \times 10^8 \; m\)

Hence \( A = 4\pi r^2 = 6 \times 10^{18} \;m^2\)

Hence \( P = 3.9 \times 10^{26}\;W\)

Now distance of Earth from \(Sun = 1.5 \times 10^{11} \; m\)

so the power per unit area at the Earth will be:

\(I = \frac{P}{4\pi r^2}=\frac{3.9 \times 10^{26} }{4\pi(1.5 \times 10^{11})^2}\)

\(=1400 \; W\;m^{-2}\)