Question

The thrust $F$ is exerted on area $2A$ and thrust $3F$ is exerted on area $\frac{A}{2}$. Find the ratio of pressure exerted.

• A. $1:2$
• B. $1:3$
• C. $1:9$
• D. $1:12$

Answer 1

The correct option is D $1:12$

Thrust per unit area is called pressure.

$P_1=\frac{F}{2A}$

$P_2=\frac{3F}{\frac{A}{2}}$

$\frac{P_1}{P_2}=\frac{\frac{F}{2A}}{\frac{3F}{\frac{A}{2}}}$
$=\frac{F}{2A}\times \frac{\frac{A}{2}}{3F}=\frac{1}{12}$

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