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Question

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±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 $$164 \pm 3 cm^2$$
Error in product of quantities: Suppose $$x=a \times b$$
Let $$\Delta a$$=absolute error in measurement of $$a,$$
$$\Delta b$$=absolute error in measurement of $$b,$$
$$\Delta x$$=absolute error in calculation of x, i.e. product of a and b.
The maximum fractional error in x is $$\dfrac{\Delta x}{x}= \pm (\dfrac{\Delta a}{a}+\dfrac{\Delta b }{b})$$
Percentage error in the value of x=(Percentage error in value of a)+(Percentage error in value of b)
According to the problem, length $$l=(16.2 \pm 0.1)cm$$

Breadth $$b=(10.1 \pm 0.1)cm$$

Area $$A=l \times b=(16.2 cm)\times (10.1 cm)=163.62 cm^2$$
As per the rule area will have only three significant figures and error will have only one significant figure.Rounding off we get,area $$ A=164 cm^2$$
If $$\Delta A$$ is error in the area, then relative error is calculated as $$\dfrac{\delta 4}{A}$$.

$$\dfrac{\Delta 4}{A}=\dfrac{\Delta l}{l}+\dfrac{\Delta b}{b}=\dfrac{0.1 cm }{16.2cm}+\dfrac{0.1 cm}{10.1 cm}$$

$$=\dfrac{1.01 +1.62}{16.2 \times 10.1}=\dfrac{2.63}{163.62}$$

$$ \Rightarrow \Delta A= A \times \dfrac{2.63}{163.62}cm^2=162.62\times \dfrac{2.63}{163.62}=2.63 cm^2$$
$$\Delta A=3 cm^2$$ (By rounding off to one significant figure)

Area, $$A=A \pm \Delta A(164 \pm 3)cm^2$$

Physics

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