Force and area are measured as 20 N and $5 m^{2}$ with errors $0.05 N$ and $0.0125 m^{2}$ . The maximum error in pressure is (SI unit)

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$4 \pm 0.02$

$4 \pm 0.05$

$4 \pm 0.0625$

$4 \pm 0.125$

answer is D.

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Detailed Solution

$P = \frac{F}{A} = \frac{20}{5} = 4 p a \Rightarrow \frac{Δ P}{P} = \frac{Δ F}{F} + \frac{Δ A}{A} \Rightarrow Δ P = P (\frac{Δ F}{F} + \frac{Δ A}{A}) = \frac{20}{5} (\frac{0.05}{20} + \frac{0.0125}{5}) = 4 (0.0025) = 0.02$

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