What is the Relationship Between Uncertainty in Diameter and Radius?

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The uncertainty in radius is half of the uncertainty in diameter, as confirmed by the formula radius = diameter / 2. If the diameter's uncertainty is ±0.02 cm, the radius uncertainty is ±0.01 cm. When dealing with percent uncertainties, the percent uncertainty remains the same for both radius and diameter. To calculate the uncertainty in the volume of a sphere, the formula used is ΔV/V = 3 × Δr/r. This approach indicates that the volume's uncertainty will be larger due to the cubic relationship with radius.
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Homework Statement


If the uncertainty in a measurement of dameter is +-0.02cm is the uncertainty in radius the same or half of this value. The formula given in wikipedia implies that it would be half of the value.

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Homework Equations





The Attempt at a Solution

 
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swain1 said:

Homework Statement


If the uncertainty in a measurement of dameter is +-0.02cm is the uncertainty in radius the same or half of this value. The formula given in wikipedia implies that it would be half of the value.

Cheers


Homework Equations





The Attempt at a Solution


Yes, it will be half of this value. If the uncertainty had been given in percent form, then the percent uncertainty on the radius and on the diameter would be the same.
 
you should had this formula in your relevant equations thingy:
radius = diameter /2

2 is a number, and has uncertainty of 0
so, use division rule
 
Ok thanks guys.
I now have to work out the uncertainty in the voume of a sphere using the radius I have. Any clues on how to get the uncertainty in this value? I would expect that it would be bigger but I can't really understand why?
 
I have this formula

deltav/v=3xdeltar/r

Is this correct?
 
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?

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