Question

a rectangular plate has length (2+-0.02)CM and width (1+-0.01)CM the maximum percentage error in the measurement of it's area​

Answer 1

Given :

• l = ( 2 ± 0.02 )
• b = ( 1 ± 0.01 )

To find :

• Maximum percentage error in area

Solution :

Area of the rectangle is given by the formula ,

A = l x b

Hence ,

$\leadsto \dfrac{\Delta A}{A} = \dfrac{\Delta l}{l} + \dfrac{\Delta b}{b}$

Now let's substitute the values in the above equation ,

$\leadsto \dfrac{\Delta A}{A} = \pm \bigg(\dfrac{0.02}{2} + \dfrac{0.01}{1}\bigg)$

$\leadsto \dfrac{\Delta A}{A} = \pm(0.01 + 0.01 )$

$\leadsto \dfrac{\Delta A}{A} = \pm 0.02$

Percentage error in area

$\leadsto \% \:error = \dfrac{\Delta A}{A} \times 100$

$\leadsto \% \:error = 0.02 \times 100$

$\leadsto \boxed{\red{ \% \:error = 2 \%}}$

The maximum percentage error in the measurement of area is 2 %.