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Earth as a System: Energy, Matter, and Life: CBSE Class 9 Physics practice questions

Practise 50 questions on Earth as a System: Energy, Matter, and Life, free, with an explanation for each answer.

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Sample questions

UV rays can damage the skin and eyes, but they are also useful. Which of these is a use of UV rays?

Ultraviolet rays from the Sun meeting a layer high in the sky
  1. Providing the energy plants use for photosynthesis
  2. Carrying radio and television signals
  3. Killing germs in water purifiers
  4. Keeping the Earth warm at night
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C. Killing germs in water purifiers

UV rays (100100–400 nm400\ \text{nm}, where 1 nm=10−9 m1\ \text{nm} = 10^{-9}\ \text{m}) carry more energy than visible light. They are used to kill germs in water purifiers and help power fluorescent lights. Because long exposure harms the eyes and skin and raises cancer risk, people use UV-protective glasses and sunscreen. Photosynthesis uses visible light, signals use radio waves, and night-time warmth is infrared trapped by greenhouse gases.

Sunlight crosses about 1.5×1011 m1.5 \times 10^{11}\ \text{m} of empty space to reach the Earth. Light travels through a vacuum at 3×108 m s−13 \times 10^{8}\ \text{m s}^{-1}. How long does the journey take?

A pulse of light travelling from the Sun to the Earth
  1. 4.5×1019 s4.5 \times 10^{19}\ \text{s}
  2. 500 s500\ \text{s}
  3. 2×10−3 s2 \times 10^{-3}\ \text{s}
  4. 50 s50\ \text{s}
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B. 500 s500\ \text{s}

t=dc=1.5×10113×108=0.5×103=500 st = \dfrac{d}{c} = \dfrac{1.5 \times 10^{11}}{3 \times 10^{8}} = 0.5 \times 10^{3} = 500\ \text{s} which is about 8.3 minutes. Light can cross a vacuum because it is an electromagnetic wave, while sound (a mechanical wave) cannot. 4.5×10194.5 \times 10^{19} s comes from multiplying instead of dividing, 2×10−32 \times 10^{-3} s from dividing the wrong way round, and 5050 s from slipping a power of ten.

At the top of the atmosphere, the solar constant is about 1400 W m−21400\ \text{W m}^{-2}. On a clear day, the highest insolation (sunlight energy received) at the ground is about 1000 W m−21000\ \text{W m}^{-2}. Roughly what percentage of the energy is lost on the way down?

Sunlight entering the atmosphere and meeting clouds, air and the ground
  1. 40%
  2. 71%
  3. 29%
  4. 0.4%
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C. 29%

Loss =1400−1000=400 W m−2= 1400 - 1000 = 400\ \text{W m}^{-2}, and 4001400×100≈29%\dfrac{400}{1400} \times 100 \approx 29\% This energy is absorbed and scattered by gases, clouds and dust. 40% divides the loss by 1000 instead of by the starting value, 71% is the fraction that does reach the ground, and 0.4% writes the 0.4 kW0.4\ \text{kW} loss as if it were a percentage. The solar constant (1.4 kW m−21.4\ \text{kW m}^{-2}) is the energy per second per square metre facing the Sun before any absorption.

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