krit.club logo

Pressure, Winds, Storms, and Cyclones - Storms, Thunderstorms, and Lightning

Grade 8CBSE

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

🔑Concepts

•

A thunderstorm is a storm with lightning and thunder, typically produced by a cumulonimbus cloud, usually accompanied by gusty winds, heavy rain, and sometimes hail.

•

Thunderstorms develop in hot, humid tropical areas like India. The rising temperatures produce strong upward rising winds which carry water droplets upwards. These droplets freeze and fall down again. The swift movement of the falling water droplets along with the rising air creates lightning and sound.

•

Lightning is a massive electric discharge between clouds, or between a cloud and the ground, caused by the accumulation of static electric charges.

•

The center of a cyclone is a calm area called the 'eye' of the storm. A large cyclone is a violently rotating mass of air in the atmosphere, 1010 to 15 km15\text{ km} high.

•

Increased wind speed is indeed accompanied by a reduced air pressure. This principle explains why high-speed winds can lift the roofs of buildings if they are weak.

•

Air moves from the region where the air pressure is high to the region where the pressure is low. The greater the difference in pressure, the faster the air moves.

•

During a thunderstorm, it is unsafe to take shelter under an isolated tall tree or near metal objects. If in a forest, take shelter under a small tree. If in open ground, do not lie on the ground; instead, squat low on the ground.

📐Formulae

P=FAP = \frac{F}{A}

Speed=DistanceTimeSpeed = \frac{Distance}{Time}

Pressure∝1Wind Speed\text{Pressure} \propto \frac{1}{\text{Wind Speed}}

💡Examples

Problem 1:

Calculate the pressure exerted by a wind that applies a force of 500 N500\text{ N} on a wall with an area of 25 m225\text{ m}^2.

Solution:

Given: Force F=500 NF = 500\text{ N} Area A=25 m2A = 25\text{ m}^2

Using the formula: P=FAP = \frac{F}{A} P=50025P = \frac{500}{25} P=20 N/m2P = 20\text{ N/m}^2

Explanation:

Pressure is the force acting per unit area. In this case, the wind exerts 2020 Pascals (Pa) of pressure on the wall.

Problem 2:

During a storm, the wind travels 180 km180\text{ km} in 33 hours. Calculate the speed of the wind in km/h\text{km/h} and m/s\text{m/s}.

Solution:

  1. Speed in km/h\text{km/h}: Speed=180 km3 h=60 km/hSpeed = \frac{180\text{ km}}{3\text{ h}} = 60\text{ km/h}

  2. To convert to m/s\text{m/s}: 60 km/h=60×1000 m3600 s60\text{ km/h} = 60 \times \frac{1000\text{ m}}{3600\text{ s}} 60 km/h=60×51860\text{ km/h} = 60 \times \frac{5}{18} 60 km/h=30018≈16.67 m/s60\text{ km/h} = \frac{300}{18} \approx 16.67\text{ m/s}

Explanation:

Wind speed is the distance covered by the wind per unit of time. Conversion from km/h\text{km/h} to m/s\text{m/s} is done by multiplying with 518\frac{5}{18}.

Problem 3:

A high-speed wind blowing over a tin roof creates a pressure difference. If the atmospheric pressure inside the house is 101325 Pa101325\text{ Pa} and the pressure outside (due to wind) drops to 100000 Pa100000\text{ Pa}, find the net pressure difference acting upwards on the roof.

Solution:

Internal Pressure (PinP_{in}) = 101325 Pa101325\text{ Pa} External Pressure (PoutP_{out}) = 100000 Pa100000\text{ Pa}

101325−1000001325\begin{array}{r} 101325 \\ -100000 \\ \hline 1325 \end{array}

ΔP=Pin−Pout=1325 Pa\Delta P = P_{in} - P_{out} = 1325\text{ Pa}

Explanation:

Because high-speed wind reduces the pressure above the roof, the higher pressure from inside the house exerts an upward force, which can lift and blow away the roof.