The head of a pin is a square of a side of 10 mm.

E

Editor

Jun 23, 2022

The head of a pin is a square of a side of 10 mm. Find the pressure on it due to a force of 20 N. $(2 \times 10^{5} Nm^{-2})$

Solution:        Force = F = 20 N                      Area of head of a pin A = $0mm \times 10mm$=  $frac{10}{10} cm \times \frac{10}{10}$cm                          = $\frac{1}{100} m \times \frac{1}{100}$ m                         = $10^{-4} m^{2}$                        Pressure under the thumb = P =?                        P =$\frac{F}{A}$                         P  =$\frac{20}{(1 \times 10^{-4} )}$ =  $2 \times 10^{5} Nm^{-2}$