Respuesta :
Answer:
The uncertainty in the volume of the sphere is [tex]1.059\times 10^{4} m^{3}[/tex]
Solution:
As per the question:
Measured radius of the sphere, R = 135.4 m
Uncertainty in the radius, [tex]\Delta R = 4.6 cm = 4.6\times 10^{- 2} = 0.046 m[/tex]
We know the volume of the sphere is:
[tex]V_{s} = \frac{4}{3}\pi R^{3}[/tex]
We know that the fractional error for the given sphere is given by:
[tex]\frac{\Delta V_{s}}{V_{s}} = \frac{4}{3}\pi.\frac{|Delta R}{R}[/tex]
where
[tex]\Delta V_{s}[/tex] = uncertainty in volume of sphere
Now,
[tex]\Delta V_{s} = \frac{4}{3}\pi 3R^{2}\Delta R[/tex]
Now, substituting the suitable values:
[tex]\Delta V_{s} = 4\pi (135.4)^{2}\times 0.046 = 1.059\times 10^{4} m^{3}[/tex]
