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Pressure, Winds, Storms, and Cyclones - High-Speed Winds Result in Lowering Air Pressure

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Air exerts pressure on all objects and in all directions due to the constant motion of its molecules.

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High-speed winds are accompanied by reduced air pressure. This is a fundamental principle where an increase in the speed of a fluid (like air) results in a decrease in pressure.

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Air moves from a region where the air pressure is high to a region where the pressure is low (Phigh→PlowP_{high} \rightarrow P_{low}).

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The greater the difference in pressure between two regions, the faster the air moves (higher wind speed).

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On heating, air expands and occupies more space. When the same amount of air occupies more space, it becomes less dense and lighter.

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Warm air rises because it is lighter than cold air. This rising of warm air creates a region of low pressure (PlowP_{low}), causing cooler, denser air to rush in to fill the gap.

📐Formulae

Pressure(P)=Force(F)Area(A)Pressure (P) = \frac{Force (F)}{Area (A)}

Density(ρ)=Mass(m)Volume(V)Density (\rho) = \frac{Mass (m)}{Volume (V)}

Wind Speed∝ΔPressureWind \, Speed \propto \Delta Pressure

💡Examples

Problem 1:

During a high-speed wind storm, why do the roofs of some weak huts or tin sheds get blown off without the walls collapsing?

Solution:

High-speed winds passing over the roof create a region of low pressure (PtopP_{top}) above the roof. The air pressure inside the hut (PinsideP_{inside}) remains high. This creates a pressure difference: ΔP=Pinside−Ptop\Delta P = P_{inside} - P_{top} If the roof is weak, this upward force lifts it off.

Explanation:

According to the principle that high-speed winds result in lower pressure, the velocity of air above the roof reduces the pressure there. Since the pressure below the roof is atmospheric and higher, it exerts an upward force that can lift the roof if it is not tied down strongly.

Problem 2:

What happens when you blow air into the space between two inflated balloons hung 8−10 cm8-10 \text{ cm} apart?

Solution:

The balloons move towards each other. This is because the high-speed air blown between them reduces the air pressure (PmiddleP_{middle}) in that space. The pressure on the outer sides (PouterP_{outer}) is higher and pushes the balloons together: Pouter>PmiddleP_{outer} > P_{middle}

Explanation:

Blowing air increases the wind speed between the balloons, which lowers the air pressure in that specific area. The higher atmospheric pressure acting from the outside of the balloons forces them inward.

Problem 3:

Calculate the pressure exerted by a wind force of 500 N500 \text{ N} acting on a wall surface of 25 m225 \text{ m}^2.

Solution:

Using the formula P=FAP = \frac{F}{A}, we have: F=500 NA=25 m2P=50025=20 N/m2\begin{array}{r} F = 500 \text{ N} \\ A = 25 \text{ m}^2 \\ \hline P = \frac{500}{25} = 20 \text{ N/m}^2 \end{array}

Explanation:

Pressure is defined as force per unit area. By dividing the total force of the wind by the area of the wall, we find the pressure exerted in Pascals (or N/m2\text{N/m}^2).