A flat head drill bit is used to drill holes of small depth. The drill bit can drill holes with a diameter size of (40 ± 0.5)mm. The depth of the hole has an uncertainty of ±1mm.
If the total percentage uncertainty of the volume made by the hole is less than 10%, what is the minimum depth, in mm, which the drill bit can make?
My working:
V=π(r^2)d => d= 4V/(πD^2)
[V=Volume; r=radius; d=depth; D=Diameter]
△d/d = △v/v + 2(△D/D)
1/d = 10/100 + 2(0.5/40)
d=8
But the options given are 15, 14, 13 and 12. What am I doing wrong?
1 answer