Pressure, Winds, Storms, and Cyclones | Class 8 Science Notes
By ConceptScroll Team · Published on 17 July 2026 · 6 min read
Pressure, Winds, Storms, and Cyclones – this guide gives you a concise, exam-ready overview of Pressure, Winds, Storms, and Cyclones from Class 8 Science, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
Pressure Exerted by Air
Air, like liquids, exerts pressure on objects around us. The layer of air surrounding the Earth is called the atmosphere, composed mainly of nitrogen, oxygen, argon, and carbon dioxide. Atmospheric pressure is the force exerted by the weight of air molecules on surfaces.
An activity involving an inverted paper plate covered with chart paper sheets of different areas demonstrates air pressure. When covered with a folded (smaller area) sheet, the plate is easier to lift compared to when covered with an unfolded (larger area) sheet. This shows that air exerts force over the area of the sheet, and the force increases with area, indicating pressure.
Inflating a balloon also demonstrates air pressure. Air inside the balloon exerts pressure on its walls, causing it to expand in all directions. If the balloon's mouth is open, air escapes from high pressure inside to lower pressure outside.
A rubber sucker pressed against a smooth surface sticks due to atmospheric pressure. Pressing the sucker expels air from between it and the surface, reducing pressure inside. The higher atmospheric pressure outside pushes the sucker against the surface, making it stick. Pulling it off requires force to overcome this pressure difference.
Atmospheric pressure is very large; for example, the force exerted by air on an area of 15 cm × 15 cm is about 2250 N, equivalent to the weight of a 225 kg object. We are not crushed because the pressure inside our bodies balances the outside atmospheric pressure.
The practical units for measuring air pressure include millibar (mb) and hectopascal (hPa), where 1 mb = 100 Pa and 1 hPa = 100 Pa.
In summary, air exerts pressure in all directions, and this atmospheric pressure plays a vital role in various natural phenomena and daily life.
📊 Diagram: Fig. 6.9a-c show an inverted paper plate with folded and unfolded chart paper sheets, demonstrating air pressure effects. Fig. 6.10 shows a girl blowing a balloon, illustrating air pressure inside the balloon. Fig. 6.11 depicts a rubber sucker sticking to a surface due to atmospheric pressure.
🧪 Activity: Activity 6.3 involves lifting an inverted paper plate covered with folded and unfolded chart papers to observe air pressure effects. Activity 6.4 uses a rubber sucker pressed on a smooth surface to demonstrate atmospheric pressure.
🔗 Connection: Understanding atmospheric pressure leads to the study of wind formation caused by pressure differences in the air.
Frequently asked questions
1. Choose the correct statement. (i) Look at Fig. 6.21 carefully. Vessel R is filled with water. When pouring of water is stopped, the level of water will be (a) the highest in vessel P (b) the highest in vessel Q (c) the highest in vessel R (d) equal in all three vessels (ii) A rubber sucker (M) is pressed on a flat smooth surface and an identical sucker (N) is pressed on a rough surface: (a) Both M and N will stick to their surfaces. (b) Both M and N will not stick to their surfaces. (c) M will stick but N will not stick. (d) M will not stick but N will stick. (iii) A water tank is placed on the roof of a building at a height 'H'. To get water with more pressure on the ground floor, one has to (a) increase the height 'H' at which the tank is placed. (b) decrease the height 'H' at which the tank is placed. (c) replace the tank with another tank of the same height that can hold more water. (d) replace the tank with another tank of the same height that can hold less water. (iv) Two vessels, A and B contain water up to the same level as shown in Fig. 6.22. PA and PB is the pressure at the bottom of the vessels. FA and FB is the force exerted by the water at the bottom of the vessels A and B. (a) PA = PB, FA = FB (b) PA = PB, FA < FB (c) PA < PB, FA = FB (d) PA > PB, FA > FB
1.(i) (d) equal in all three vessels. Explanation: The water level will be equal in all connected vessels due to the principle of communicating vessels.
(ii) (c) M will stick but N will not stick. Explanation: A rubber sucker sticks to a smooth surface because of air pressure difference; on a rough surface, air can enter, so it won't stick.
(iii) (a) increase the height 'H' at which the tank is placed. Explanation: Pressure due to water depends on height of water column, so increasing height i
2. State whether the following statements are True [T] or False [F]. (i) Air flows from a region of higher pressure to a region of lower pressure. (ii) Liquids exert pressure only at the bottom of a container. (iii) Weather is stormy at the eye of a cyclone. (iv) During a thunderstorm, it is safer to be in a car.
(i) True. Air moves from high to low pressure causing wind. (ii) False. Liquids exert pressure in all directions, not only at the bottom. (iii) False. The eye of a cyclone is calm with low pressure. (iv) True. Being inside a car is safer during thunderstorms as it acts like a Faraday cage.
3. Fig. 6.23a shows a boy lying horizontally, and Fig. 6.23b shows the boy standing vertically on a loose sand bed. In which case does the boy sink more in sand? Give reasons.
The boy sinks more when standing vertically (Fig. 6.23b). Reason: Pressure is force per unit area. When standing, the boy's weight is concentrated on a smaller area (feet), increasing pressure on sand, causing him to sink more. When lying horizontally, the weight is distributed over a larger area, reducing pressure and sinking.
4. An elephant stands on four feet. If the area covered by one foot is 0.25 m², calculate the pressure exerted by the elephant on the ground if its weight is 20000 N.
Given: Weight, W = 20000 N Area per foot, A_foot = 0.25 m² Number of feet = 4 Total area, A_total = 4 × 0.25 = 1 m²
Pressure, P = Force / Area = W / A_total = 20000 N / 1 m² = 20000 Pa
Therefore, the pressure exerted by the elephant on the ground is 20000 pascals (Pa).
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