A brick with dimensions of 20 cm x 10 cm x 5 cm has a weight of 500 g Calculate the pressure exerted by it when it rests on different faces. In which case is the pressure exerted is the greatest
Answers
Therefore the maximum pressure is exerted on the 10 cm × 5 cm face.
Given:
The dimensions of the brick = 20 cm × 10 cm × 5 cm
Weight exerted = 500g N
Where g = Acceleration due to gravity = 9.81 m /
To Find:
The case in which the pressure exerted is maximum.
Solution:
The given question can be solved as shown below.
Three different surface areas can be possible as listed below:
Case-1: 1st Surface Area = SA₁ = 20 cm × 10 cm = 200 cm²
Case-2: 2nd Surface Area = SA₂ = 10 cm × 5 cm = 50 cm²
Case-3: 3rd Surface Area = SA₃ = 20 cm × 5 cm = 100 cm²
We know that, Pressure exerted = ( Force ) / ( Surface Area )
As weight is constant in all the cases,
⇒ Pressure exerted ∝ 1/ ( Surface Area )
Hence the maximum pressure is exerted on the face with minimum Surface Area.
The minimum Surface Area is obtained in the 2nd case ( SA₂ ). Hence Pressure exerted is maximum on the 2nd Surface Area ( Case-2 )
The value of maximum pressure exerted = ( Weight ) / ( SA₂ )
⇒ Maximum pressure exerted on face 2 = ( 500g ) / ( 50 ) = 10g N/cm²
Therefore the maximum pressure is exerted on the 10 cm × 5 cm face.
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