if a 200 newton force is acting in the block of dimensions 20 cm x 10 cm x 5 cm , find the maximum and minimum pressure exerted by the sides of the block
Answers
Maximum pressure exerted by the force is 40000 N/m²
Minimum pressure exerted by the force is 10000 N/m²
Explanation:
Given:
Dimensions of the block = 20 cm × 10 cm × 5 cm
And a 200 N force is applied
To find out:
The maximum and minimum pressure exerted by the force on the sides of the block
Solution:
There will be 6 surfaces of the block
With two surface each of area = 20 cm x 10 cm = 200 cm²
Other two surfaces each of area = 10 cm x 5 cm = 50 cm²
Last two surfaces each of area = 20 cm x 5 cm = 100 cm²
As we know that
i.e. pressure is inversely proportional to the area
Thus,
The pressure will be maximum when area is minimum
And, the pressure will be minimum when the area is maximum
Here the minimum area is 50 cm² and the maximum area is 200 cm²
Therefore,
The maximum pressure is
N/m²
N/m²
N/m²
The minimum pressure is
N/m²
N/m²
N/m²
Hope this answer is helpful.
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Given:
Newton force = 200N
Dimensions of block = 20 cm x 10 cm x 5 cm
To Find:
Maximum and minimum pressure exerted by the sides of the block
Solution:
Let the pressures exerted by the brick while resting on different faces be = P1, P2 and P3.
Case (i) : When the block is resting on 20cm × 10cm face.
F = 200 N
= 20 /200 x 10 / 200
= .005 m²
Pressure = Force /Area
= 200 / .005
= 40,000 pa
Case (ii) : When the block is resting on 20cm × 5cm face
F = 200 N
Area = 20 cm x 5cm
20 /200 m x 5/ 200 m
= .0025 m²
Pressure = Force /Area
= 200 / .0025
= 80,000 pa
Case (iii) : When the block on 10cm×5cmface
F = 200 N
= 10 /200 x 5 / 200
= .00125 m²
Pressure = Force /Area
= 200 / .00125
= 1,60,000 pa
Answer: the maximum pressure is 1,60,000 pa and the minimum pressure is 40,000 pa