Question

A force of(2500+/-5)N is applied over an area of (0.32 +/- 0.02)m² .Calculate the pressure exerted over the area with error limits.

Answer 1

$$\begin{gathered} \mathrm{Pressure}=\frac{\mathrm{Force}}{\mathrm{Area}} \\ \mathrm{P}=\mathrm{F}/\mathrm{A},\mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{this}\mathrm{eqn}\mathrm{and}\mathrm{finding}\mathrm{differentiation}, \\ \frac{∆\mathrm{P}}{\mathrm{P}}=\frac{∆\mathrm{F}}{\mathrm{F}}+\frac{∆\mathrm{A}}{\mathrm{A}},\mathrm{but}\mathrm{we}\mathrm{are}\mathrm{given}∆\mathrm{F}=\pm 5\mathrm{N},∆\mathrm{A}=\pm 0.02{\mathrm{m}}^2\mathrm{and} \\ \mathrm{P}=\mathrm{F}\times \mathrm{A}=2500\times 0.32=800. \\ \mathrm{Therefore},\frac{∆\mathrm{P}}{800}=\frac{5}{2500}+\frac{0.02}{0.32}=0.002+0.0625=0.0645. \\ ∆\mathrm{P}=800\times 0.0645=52.60. \\ \mathrm{Therefore}\mathrm{P}+∆\mathrm{P}=(800\pm 51.60)\mathrm{N}/{\mathrm{m}}^2. \end{gathered}$$