Question

A pressure of 1.5 pascal acts on a surface of area 13 square centimetres. The actual thrust acting on the surface is newton.

• A. $7.2\times {10}^{-4}$
• B. $11.3\times {10}^{-4}$
• C. $19.5\times {10}^{-4}$

Answer 1

The correct option is C $19.5\times {10}^{-4}$

Given: Area of contact, $A=13c{m}^2=13\times {10}^{-4}{m}^2$
Let the thrust be $F$.
Pressure acting on the surface, $P=1.5Pa=\frac{F}{A}$
$1Pa=1N{m}^{-2}$
$⟹1.5Pa=\frac{F}{13\times {10}^{-4}}$
$⟹F=1.5N{m}^{-2}\times 13\times {10}^{-4}{m}^2$
$⟹F=19.5\times {10}^{-4}N$