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Question

The length and breadth of a rectangular sheet are 16.2 cm and 10.1cm, respectively. The area of the sheet inappropriate significant figures and error is:

A
164±3cm2
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B
163.62±2.6cm2
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C
163.6±2.6cm2
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D
163.62±3cm2
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Solution

The correct option is D 164±3cm2
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 Δxx=±(Δaa+Δbb)
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 δ4A.

Δ4A=Δll+Δbb=0.1cm16.2cm+0.1cm10.1cm

=1.01+1.6216.2×10.1=2.63163.62

ΔA=A×2.63163.62cm2=162.62×2.63163.62=2.63cm2
ΔA=3cm2 (By rounding off to one significant figure)

Area, A=A±ΔA(164±3)cm2

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