First slide

Accuracy; precision of instruments and errors in measurements

Question

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:

Moderate

A

$(0.34 \pm 0.02) cm$
B

$(0.17 \pm 0.02) cm$
C

$(0.17 \pm 0.01) cm$
D

$(0.34 \pm 0.01) cm$

Solution

$\begin{array}{l} D = (4 .23 \pm 0 .01) cm \\ d = (3 .89 \pm 0 .01) cm \\ Δt = \frac{(D - d)}{2} \\ = \frac{(4 .23 \pm 0 .01) - (3 .89 \pm 0 .01)}{2} \\ = \frac{(4 .23 - 3 .89) \pm (0 .01 + 0 .01)}{2} \\ = \frac{(0 .34 \pm 0 .02)}{2} cm \\ = (0 .17 \pm 0 .01) cm \end{array}$

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