The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm respectively. The area of the sheet inappropriate significant figures and error is: a) 164 ± 3 cm2 b) 163.62 ± 2.6 cm2 c) 163.6 ± 2.6 cm2 d) 163.62 ± 3 cm2
Error in product of quantities: Suppose x=a×b Let Δa=absolute error in measurement of a, Δb=absolute error in measurement of b, Δx=absolute error in calculation of x, i.e. product of a and b. The maximum fractional error in x is xΔx=±(aΔa+bΔb) Percentage error in the value of x=(Percentage error in value of a)+(Percentage error in value of b) According to the problem, length l=(16.2±0.1)cm Breadth b=(10.1±0.1)cm Area A=l×b=(16.2cm)×(10.1cm)=163.62cm2 As per the rule area will have only three significant figures and error will have only one significant figure.Rounding off we get,area A=164cm2 If ΔA is error in the area, then relative error is calculated as Aδ4. AΔ4=lΔl+bΔb=16.2cm0.1cm+10.1cm0.1cm =16.2×10.11.01+1.62=163.622.63 ⇒ΔA=A×163.622.63cm2=162.62×163.622.63=2.63cm2 ΔA=3cm2 (By rounding off to one significant figure) Area, A=A±ΔA(164±3)cm2
