The external and internal diameter of a tube are measured as (32 ± 2)mm and (21 ± 1)mm respectively. Determine the percentage error in the thickness of the tube.
27%
14%
9%
3%
Explanation
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Discussions (27)
myschool pls make correction on this
the answer is 27%
Reference: revision physics page 3
To find the percentage error in the thickness of the tube:
Thickness = (External diameter - Internal diameter) / 2
= (32 - 21) / 2
= 5.5 mm
Error in thickness = (Error in external diameter + Error in internal diameter) / 2
= (2 + 1) / 2
= 3 / 2
= 1.5 mm
Percentage error = (Error in thickness / Thickness) × 100
= (1.5 / 5.5) × 100
= 27.27% ≈ 27%
The final answer is: 27%
27
External diameter- internal=11
External error+internal error= 3
3/11×100/1=27%
First of all the answer is 27%.. You don't subtract error only the actual value can be subtracted.. And u didn't even find the radius for both the external and internal if u find it and subtract the external and internal what ever you get will be used to find errors percentages
Myschool, please correct this. The Newer is option A.
Error in thickness/thickness x 100/1
Error in thickness = 2 + 1 /2
=1.5
Thickness = 32/2 - 21/2
16 - 10.5 = 5.5
1.5/5.5 x 100/1
=27%
The answer is supposed to be 27%, please correction is needed.
you used 34 and 22 to solve the question. check well
