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

Options

A. `164+-3cm^(2)`
B. `163.62+-2.6cm^(2)`
C. `163.6+-2.6cm^(2)`
D. `163.62+-3cm^(2)`

Answer

A

Explanation

If `trianglex` is error in a physical quantity, then relative error is calculated as `(trianglex)/(x)` Given, length`l=(16.2+-0.10)` cm
Breadth`b=(10.1+-0.1)` cm
Area `A=lxxb`
`=(162cm)xx(10.1cm)=163.62cm^2`
Rounding off to three significant digits, area `A=164cm^2`
Rounding off to three significant digits, area `A=164cm^2`
`(triangleA)/(A)=(trianglel)/(l)+(triangleb)/(b)=(0.1)/(16.2)+(0.1)/(10.1)`
`=(1.01+1.62)/(16.2xx10.1)=(2.63)/(163.62)`
`impliestriangleA=Axx(2.63)/(163.62)`
`=163.62xx(2.63)/(163.62)=2.63cm^2`
`triangleA=3cm^2` (By rounding off to one significant figure)
Area, `A=A+-triangleA=(164+-3)cm^2`