The internal and external diameters of a holl | vedclass

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(D) Given:Internal diameter $d = (5.73 \pm 0.01) 	ext{ cm}$External diameter $D = (6.01 \pm 0.01) 	ext{ cm}$The thickness $t$ of the cylinder wall i

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$(0.14 \pm 0.01) 	ext{ cm}$

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(D) Given:Internal diameter $d = (5.73 \pm 0.01) 	ext{ cm}$External diameter $D = (6.01 \pm 0.01) 	ext{ cm}$The thickness $t$ of the cylinder wall is given by the formula:$t = \frac{D - d}{2}$First,calculate the mean value:$t_{	ext{mean}} = \frac{6.01 - 5.73}{2} = \frac{0.28}{2} = 0.14 	ext{ cm}$Next,calculate the uncertainty in thickness:Since the thickness involves subtraction,the absolute errors add up:$\Delta t = \frac{\Delta D + \Delta d}{2} = \frac{0.01 + 0.01}{2} = \frac{0.02}{2} = 0.01 	ext{ cm}$Therefore,the thickness is $(0.14 \pm 0.01) 	ext{ cm}$.

The internal and external diameters of a hollow cylinder measured with vernier calipers are $(5.73 \pm 0.01) \text{ cm}$ and $(6.01 \pm 0.01) \text{ cm}$ respectively. Then the thickness of the cylinder wall is

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