Physics, asked by kanhaiya4036, 1 year ago

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

Answers

Answered by AR17
9
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.

nalinsingh: Superb!!
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