Question

Thrust of 'f' N is exerted on an area 2a & thrust of '3f' N is exerted on an area a12 .find the ratio of pressure exerted

Answer 1

Pressure = f/2a
Pressure = 3f/12a=f/4a
Ratio of pressure =f/2a/f/4a=4/2
=2:1

Answer 2

Answer:

this is the correct ans

Explanation:

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