Question

Thrust of 'F'N is exerted on an area of 2A and thrust of 3F is exerted on an area of A/2. Find the ratios of their pressure

Answer 1

Hey, there!

Here's the answer you are looking for

Pressure = Force/Area

☛ In the 1st case,

P1 = F/2A

☛ In the 2nd case,

P2 = 3F/(A/2) = 6F/A


☞ So, the ratio,
$\frac{P1}{P2} = \frac{F/2A}{6F/A} \\ \\ \: \: \: \: \: \: \: \: = \frac{F.A}{6F.2A} \: = \: \frac{1}{12}$


Thus, the ratio of pressure is 1:12.


Hope that helps!


Thanks.