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

Thickness =(4.23±0.1)−(3.89±0.1)2=(4.23−3.89)±(0.02)2=0.34±0.022=0.17±0.01

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

Thickness =(4.23±0.1)−(3.89±0.1)2=(4.23−3.89)±(0.02)2=0.34±0.022=0.17±0.01

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

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(0.17 $\pm$ 0.01) cm

(0.34 $\pm$ 0.02) cm

(0.17 $\pm$ 0.02) cm

(0.34 $\pm$ 0.01) cm

answer is C.

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Detailed Solution

Thickness $= \frac{(4.23 \pm 0.1) - (3.89 \pm 0.1)}{2}$
$\begin{array}{l} = \frac{(4.23 - 3.89) \pm (0.02)}{2} \\ = \frac{0.34 \pm 0.02}{2} = 0.17 \pm 0.01 \end{array}$

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