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Q. The external and internal radius of a hollow cylinder are measured to be $(4.23 \,\pm \,0.01)\, cm$ and $(3.89 \,\pm \, 0.01) \,cm$, respectively. The thickness of the wall of the cylinder is

Physical World, Units and Measurements

Solution:

Thickness of wall $(t) = r, - r_2$
$t = 4.23 - 3.89 = 0.34\, cm$
$\Delta t = \pm (\Delta r_1 + \Delta r_2) = (0.01 + 0.01) = 0.02\, cm$
$\therefore $ Thickness of wall $=( t \pm \Delta t) = (0.34 \pm 0.02) \,cm$