Correct statement about error and uncertainty

  • Thread starter Thread starter songoku
  • Start date Start date
  • Tags Tags
    Error Uncertainty
Click For Summary
The discussion revolves around understanding the differences between options (b) and (d) regarding error and uncertainty in measurements. The main point of contention is whether averaging multiple readings affects fractional uncertainty, with the belief that averaging can reduce random error. There is a humorous suggestion that the average could be smaller if the object being measured, like a balloon, is deflating. Participants express uncertainty about the exclusivity of correct answers, noting that typically only one answer is correct unless the question itself is flawed. The conversation highlights the complexity of interpreting error and uncertainty in measurement contexts.
songoku
Messages
2,508
Reaction score
402
Homework Statement
Which of the following statements is correct?
a. Checking for zero error in voltmeter before measuring a voltage will help to reduce random error

b. Measuring the timing of a larger number of oscillations will help to reduce the fractional uncertainty of the period

c. Plotting a graph of voltage and current readings for an ohmic device and using its gradient to find resistance will help to eliminate random error

d. In the measurement of the diameter of a sphere, taking more readings and finding the average value of these readings will help to reduce the fractional uncertainty of the diameter
Relevant Equations
Not sure
I know option (a) and (c) are wrong but I don't know the difference between option (b) and (d) because I think they are the same (taking average to get more accurate reading and reduce random error).

My guess is (b) because I think for (d) the fractional uncertainty maybe can stay the same because taking average will reduce random error (the uncertainty) and also can reduce the reading of the diameter (I mean the average of 5 readings will be smaller than average of 4 readings) so the fractional uncertainty (##\frac{\Delta ~ diameter}{diameter}##) stays the same

I am not sure whether my reasoning is correct. Thanks
 
Physics news on Phys.org
songoku said:
I mean the average of 5 readings will be smaller than average of 4 readings
No reason why it should ...
(unless the balloon is deflating :smile:)

You sure only one and not two can be correct ?
 
BvU said:
No reason why it should ...
(unless the balloon is deflating :smile:)

You sure only one and not two can be correct ?
Well I can't say I am sure but from all the questions given, usually there is only one correct answer, unless there is something wrong with the question
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
1K