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Question

The thrust F is exerted on area 2 A and thrust 3 F is exerted on area A2. Find the ratio of pressure exerted.

A
1:2
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B
1:3
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C
1:9
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D
1:12
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Solution

The correct option is D 1:12
Thrust per unit area is called pressure.

P1=F2A

P2=3FA2

P1P2=F2A3FA2
=F2A×A23F=112

Thus, the ratio of pressure exerted is 1:12.

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