The external and internal diameters of a hollow cylinder are measured to be `(4.23+-0.01)` cm and `(3.89+-0.01)`cm. The thickness of the wall of the cylinder is

Options

A. `(0.34 pm0.02) cm`
B. `(0.17pm 0.02) cm`
C. `(0.17pm0.02) cm`
D. `(0.34 pm0.01) cm`

Answer

C

Explanation

D=(4.32 `pm` 0.01) cm
d=(3.89 `pm` 0.01) cm
`Deltat=(D-d)//2`
`=((4.23pm0.01)-(3.89 pm 0.01))/2`
`=((4.23-3.89) pm(0.01-0.01))/(2)`
`=(0.34 pm 0.02)//2 cm`
`=(0.17 pm 0.01) cm`