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(D) Pressure $P$ is defined as the force $F$ per unit area $A$. For a square of side $L$,the area is $A = L^2$.Thus,$P = \frac{F}{L^2}$.The relative e

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$8$

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(D) Pressure $P$ is defined as the force $F$ per unit area $A$. For a square of side $L$,the area is $A = L^2$.Thus,$P = \frac{F}{L^2}$.The relative error in pressure is given by the formula: $\frac{\Delta P}{P} = \frac{\Delta F}{F} + 2 \frac{\Delta L}{L}$.To find the maximum percentage error,we multiply by $100 \%$:$\frac{\Delta P}{P} 	imes 100 \% = \left( \frac{\Delta F}{F} 	imes 100 \% ight) + 2 \left( \frac{\Delta L}{L} 	imes 100 \% ight)$.Given $\frac{\Delta F}{F} 	imes 100 \% = 4 \%$ and $\frac{\Delta L}{L} 	imes 100 \% = 2 \%$.Substituting these values: $\frac{\Delta P}{P} 	imes 100 \% = 4 \% + 2(2 \%) = 4 \% + 4 \% = 8 \%$.

$A$ force $F$ is applied on a square area of side $L$. If the percentage error in the measurement of $L$ is $2 \%$ and that in $F$ is $4 \%$,what is the maximum percentage error in pressure?

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