Question

A hole (diameter 5+-0,1 mm) is drilled into a cube (edge 20+- 0,2 mm). Calculate the volume of the body and its cumulative error.

Answer 1

Question #73250

A hole (diameter $5 + -0.1 \, \text{mm}$) is drilled into a cube (edge $20 + -0.2 \, \text{mm}$). Calculate the volume of the body and its cumulative error.

Answer:

The volume of the body equals to

$$V = V_{c} - V_{h} = a^{3} - a \cdot \frac{\pi d^{2}}{4},$$

where $V_{a}$ and $V_{b}$ – volume of a cube and a hole, $a$ – edge of a cube, $d$ – diameter of a hole.

The cumulative error of volume of the body we should use equals to:

$$\Delta V = \sqrt{ \left( V_{c} \frac{\Delta a \sqrt{3}}{a} \right)^{2} + V_{h}^{2} \left[ \left( \frac{\Delta a}{a} \right)^{2} + 2 \left( \frac{\Delta d}{d} \right)^{2} \right] },$$

where $\Delta a$ and $\Delta d$ – errors in measurements of $a$ and $d$.

So, the volume of the body is

$$V = 20^{3} - 20 \cdot \frac{\pi 5^{2}}{4} = 8000 - 19,63 = 7980,37 \, \text{mm}^{2}.$$

And its cumulative error is

$$\Delta V = \sqrt{ \left( 8000 \cdot \frac{0,2 \cdot \sqrt{3}}{20} \right)^{2} + 19,63^{2} \left[ \left( \frac{0,2}{20} \right)^{2} + 2 \left( \frac{0,1}{5} \right)^{2} \right] } = 138,56 \, \text{mm}^{2}.$$

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