1.

If L=20.04m pm0.01m, B=2.52 m pm 0.02 m. The value of (LxxB) is

Answer»

`(50.5pm 0.84)m^(2)`
`(50.5pm 0.42)m^(2)`
`(50.5pm0.21)m^(2)`
`(50.5pm0.50)m^(2)`

Solution :Given errors are absolute errors, while the RULE says that percentage errors are to be added. HENCE, the first STEP will be to convert the given absolute errors into percentage errors.
`L=20.04m pm (0.01)/(20.04)xx100%=20.04m pm 0.05%`
`B=2.52m pm (0.02)/(2.52)xx100%=2.52m pm 0.79%`
`LxxB=(20.04xx2.52)m^(2)pm(0.05+0.79)%=50.5m^(2)pm 0.84%`
This is the result. However, since the data given in the equation was in terms of absolute errors, so we should given our result ALSO in absolute errors.
`LxxB=50.50m^(2)pm (0.84)/(100)xx50.50m^(2)=50.50m^(2)pm0.42m^(2)`


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