CMDLearn

Learn

Class 9 Science: Earth as a System, Energy, Matter and Life, 8 practice questions with answers

CMDLearn ·

These are 8 practice questions on Chapter 13, Earth as a System: Energy, Matter, and Life, of the new NCERT Class 9 Science textbook, Exploration (First Edition, April 2026). Each has a worked answer and the textbook page it comes from. They cover the solar energy falling on a panel, how much light a surface reflects (albedo), how air cools with height in the troposphere, the planetary pressure belts, salinity and ocean gyres, how legumes add nitrogen to soil, eutrophication, and the fast and slow parts of the carbon cycle. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the numbers is taken from the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Exploration: Textbook of Science for Grade 9, Chapter 13, printed pages 252 to 268):

Energy from the Sun

Question 1. A solar water heater has a collecting panel of area 1.8 m². On a hazy afternoon the insolation on it is 0.6 kW m⁻². How much solar energy falls on the panel in 40 minutes?

Answer: 2,592,000 J, about 2.6 × 10⁶ J. Intensity = 0.6 kW m⁻² = 600 J s⁻¹ m⁻², and 40 minutes = 2400 s. E = intensity × area × time = 600 × 1.8 × 2400 = 2,592,000 J. The insolation is below the clear-sky figure of about 1 kW m⁻² because haze absorbs and scatters some of the sunlight. (Insolation, page 254; E = intensity × area × time, Example 13.1, page 255.)

Question 2. On a summer afternoon the air temperature at the ground is 33 °C. Using the textbook's rate of cooling in the troposphere, estimate the air temperature 3.4 km above the ground.

Answer: about 10.9 °C. In the troposphere the temperature falls by about 6.5 °C for every kilometre of height. Drop = 6.5 × 3.4 = 22.1 °C, so the temperature ≈ 33 − 22.1 = 10.9 °C. The height is well inside the troposphere, which is about 12 km high on average, so the rule applies. (Troposphere, page 258; Table 13.2.)

Question 3. Sunlight of 520 W m⁻² falls on a surface with an albedo of 0.35. How much of it is reflected and how much is absorbed (ignore the air)? Another surface has an albedo of 0.15. Which of the two heats up faster in the same sunlight?

Answer: 182 W m⁻² reflected and 338 W m⁻² absorbed; the surface with albedo 0.15 heats up faster. Albedo is the fraction reflected: 0.35 × 520 = 182 W m⁻². The rest, 520 − 182 = 338 W m⁻² (a fraction of 0.65), is absorbed. A surface with a lower albedo reflects less and absorbs more, so it heats up faster. (Albedo, page 256.)

Winds and ocean currents

Question 4. A weather balloon team travels from the equator to the North Pole. Name, in order, the pressure belts they cross, and say whether air rises or sinks in each. Which way are the winds deflected on this half of the Earth?

Answer: equatorial low (air rises), sub-tropical high near 30° N (air sinks), sub-polar low near 60° N (air rises), polar high (air sinks); winds are deflected to the right. Strong heating at the equator makes air rise; it cools and sinks at the sub-tropical high; part of it flows poleward and rises again at the sub-polar low; cold, dense air sinks over the pole. Because the Earth rotates, winds in the Northern Hemisphere bend to the right of their path. (Pressure belts and planetary winds, page 260.)

Question 5. Two bodies of sea water have the same temperature, but water A is less salty than water B. Which one tends to stay near the surface? And in which direction does a large ocean gyre turn in the Southern Hemisphere?

Answer: A stays near the surface; the gyre turns counter-clockwise. Water with lower salinity is less dense, so it stays near the surface, while the saltier, denser water B sinks and moves at deeper levels. The Earth's rotation turns gyres clockwise in the Northern Hemisphere and counter-clockwise in the Southern Hemisphere. (Ocean currents, page 261.)

Cycles of matter

Question 6. A farmer grows moong, a legume, on a field before sowing a cereal crop. Why does the moong crop add usable nitrogen to the soil, even though plants cannot use the nitrogen gas in the air?

Answer: because of nitrogen-fixing bacteria in its root nodules. Rhizobium living in the root nodules of legumes converts atmospheric nitrogen (N₂) into ammonia. Nitrifying bacteria then turn ammonia into nitrite and nitrite into nitrate, which plants can absorb from the soil. (Nitrogen cycle, page 264.)

Question 7. After heavy rain, fertiliser from nearby fields washes into a village pond. Within weeks the pond is covered with a thick green layer, and dead fish float on it. What is this process called, and why do the fish die?

Answer: eutrophication. The extra nitrates from the fertiliser make algae grow very fast (an algal bloom). The bloom uses up the oxygen dissolved in the water, so the fish die. (Human impact on Earth's processes (section 13.4), page 265; Fig. 13.17.)

Question 8. Sort these into the fast carbon cycle or the slow carbon cycle: (a) a neem tree taking in CO₂ in sunlight; (b) a fallen branch rotting and releasing CO₂; (c) plants buried underground turning into coal. When the coal is burnt, is its carbon returned to the air slowly or quickly?

Answer: (a) and (b) fast; (c) slow; burning returns it quickly. Photosynthesis and decomposition are part of the fast cycle, over days to years. Buried plants and animals become fossil fuels in the slow cycle, over millions of years. When fossil fuels are burnt, their carbon goes back into the air as CO₂ on a very short time scale. (Carbon cycle, page 263.)

Practise more, and check your marks

Where CMDLearn fits

CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.

Sources

All opened and read on 22 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 251.

The definitions, rules and the cooling rate are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.