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. $164\pm 3c{m}^2$
• B. $163.62\pm 2.6c{m}^2$
• C. $163.6\pm 2.6c{m}^2$
• D. $163.62\pm 3c{m}^2$

Answer 1

Answer: (A)

$164\pm 3c{m}^2$

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

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

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

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

Area, $A=A\pm \Delta A(164\pm 3)c{m}^2$