The error in the measurement of the length an | vedclass

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(B) Given: Length $l = 1 \ m$, Breadth $b = 50 \ cm = 0.5 \ m$.The relative error in length is $\frac{\delta l}{l} = 1 \% = 0.01$ and in breadth is $\

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$(0.5 \pm 0.01) \ m^2$

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(B) Given: Length $l = 1 \ m$, Breadth $b = 50 \ cm = 0.5 \ m$.The relative error in length is $\frac{\delta l}{l} = 1 \% = 0.01$ and in breadth is $\frac{\delta b}{b} = 1 \% = 0.01$.The area of the rectangle is $A = l 	imes b = 1 	imes 0.5 = 0.5 \ m^2$.The relative error in area is given by $\frac{\delta A}{A} = \frac{\delta l}{l} + \frac{\delta b}{b}$.Substituting the values: $\frac{\delta A}{A} = 0.01 + 0.01 = 0.02$.The absolute error in area is $\delta A = A 	imes 0.02 = 0.5 	imes 0.02 = 0.01 \ m^2$.Therefore, the area of the table including error is $A \pm \delta A = (0.5 \pm 0.01) \ m^2$.

The error in the measurement of the length and the breadth of a rectangular table is $1 \%$. If the length and breadth of the table are $1 \ m$ and $50 \ cm$ respectively, then the area of the table including error is

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