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. `164pm3" cm"^(2)`
B. `163.62pm2.6" cm"^(2)`
C. `163.6pm2.6" cm"^(2)`
D. `163.62pm3" cm"^(2)`

Answer

A

Explanation

Given, length `l=(16.2pm0.1)" cm"`
Breadth `b=(10.1pm0.1)" cm"`
Area `A=lxxb`
`" "`=(16.2 cm)`xx`(10.1 cm)= 163.62 `"cm"^(2)`
Rounding off to three significant digits, area `A=164" cm"^(2)`
`(DeltaA)/(A)=(Deltal)/(l)+(Deltab)/(b)=(0.1)/(10.1)`
`=(1.01+1.62)/(163.2xx10.1)=(2.63)/(163.62)`
`rArr" "DeltaA=Axx(2.63)/(163.62)=163.62xx(2.63)/(163.62)`
`" "=2.63" cm"^(2)`
`DeltaA=3" cm"^(2)` (By rounding off to one significant figure)
`therefore "Area",A=ApmDeltaA=(1654pm3)" cm"^(2)`.