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  4. 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

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Q. The external and internal diameters of a hollow cylinder are measured to be $\left(4.23\pm0.01\right)$ cm and $\left(3.89\pm0.01\right)$cm. The thickness of the wall of the cylinder is

Physical World, Units and Measurements

Solution:

$D=\left(4.23\pm0.01\right)cm$
$d=\left(3.89\pm0.01\right)cm$
$\Delta t=\left(D-d\right)/2=\left[\left(4.23\pm0.01\right)-\left(3.89\pm0.01\right)\right]/2$
$=\left[\left(4.23-3.89\right)\pm\left(0.01+0.01\right)\right]/2$
$=\left(0.34\pm0.02\right)/2\, cm=\left(0.17\pm0.01\right)cm$