The diameter of two cylingers are found to be `d_(1)= (2 pm 0.01)cm" and "d_(2)=(3.1 pm 0.03)cm`. Find the difference between both the diameters along with the error limits.

Explanation

Given `d_(1)=(2 pm 0.01)cm`
`d_(2)= (3.1 pm 0.03)cm`
Difference, `d.= d_(2)-d_(1)= 3.1-2= 1.1cm`
`triangled_(1)= pm (triangleD_(2)-triangled_(1))= pm(0.01+0.03)`
`triangled_(1)= pm 0.04`
Percentage error `=(pm 0.04)/(1.1)xx100= 3.636%`
Thus difference between diameter `=(1.1 pm 0.04)cm`
`=1.1cm pm 3.64%`.