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

The correct option is D (0.17 ± 0.01) cmExternal diameter #160; #160; d_e = (4.23± 0.01) cm So external radius #160; r_e = d_e2 = (2.12± ...

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Standard XII

Physics

Error and Uncertainty

External and ...

Question

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

(0.34 \pm 0.03)

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(0.17 \pm 0.04)

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(0.17 \pm 0.01)

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(0.34 \pm 0.02)

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Solution

The correct option is D $(0.17 \pm 0.01)$ cm
External diameter $d_{e} = (4.23 \pm 0.01) c m$
So external radius $r_{e} = \frac{d_{e}}{2} = (2.12 \pm 0.005) c m$
Internal diameter $d_{i} = (3.89 \pm 0.01) c m$
Internal diameter $r_{i} = \frac{d_{i}}{2} = (1.95 \pm 0.005) c m$
Thickness of the wall $t = r_{e} - r_{i} = (2.12 - 1.95) \pm (0.005 + 0.005) c m = (0.17 \pm 0.01) c m$

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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

The correct option is D (0.17±0.01) cmExternal diameter de=(4.23±0.01) cmSo external radius re=de2=(2.12±0.005) cmInternal diameter di=(3.89±0.01) cmInternal diameter ri=di2=(1.95±0.005) cmThickness of the wall t=re−ri=(2.12−1.95)±(0.005+0.005) cm=(0.17±0.01) cm

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