Question
A force of $(2500 \pm 5)$ N is applied over an area of $(0.32 \pm 0.02) \text{m}^2.$ Calculate the pressure exerted over the area.
✓
Answer
Here force, $\text{F}=(2500\pm5)\text{N}$ and area $\text{A}=(0.32\pm0.02)\text{m}^2$ Pressure $\text{P}=\frac{\text{F}}{\text{A}}$ $\text{P}=\frac{2500}{0.32}$ $\therefore\text{P}=7812.5\text{Nm}^{-2}$ As per the principle of error, $\Rightarrow\frac{\Delta\text{P}}{\text{P}}=\pm\Big(\frac{\Delta\text{F}}{\text{F}}+\frac{\Delta\text{A}}{\text{A}}\Big)$ $\Rightarrow\frac{\Delta\text{P}}{7812.5}=\pm\Big(\frac{5}{2500}+\frac{0.02}{0.32}\Big)$ $\Rightarrow\Delta\text{P}=\pm7812.5\Big(\frac{1}{500}+\frac{1}{16}\Big)$ $=\pm7812.5\times\frac{516}{8000}=\pm503.9\text{Nm}^{-2}$ Hence, the pressure with error limit $= (7812.5 \pm 503.9) \text{Nm}^{-2}$
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