Assume you have measured your little metal sphere's diameter to be 2.500 ± 0.005 cm and its mass to be 63.00 ± 0.05 g.

whats the volume and the error of the volume

Given the diameter of your sphere and the error on that reading (given in the introduction to this problem), what is the percent error on the measurement?

Using your simple rule devised in part B. g) of the lab, what would you estimate the percent error to be on the volume?

What percent error on the volume do you find if you use your answer to part A as the error on the volume?

In this case does the rule of thumb (part C) overestimate or underestimate the more precise error determination (part D) ?

a: Overestimate

b: Underestimate

Question

Physics

Fabian Hinton

27 August, 02:15

Assume you have measured your little metal sphere's diameter to be 2.500 ± 0.005 cm and its mass to be 63.00 ± 0.05 g.

whats the volume and the error of the volume

Given the diameter of your sphere and the error on that reading (given in the introduction to this problem), what is the percent error on the measurement?

Using your simple rule devised in part B. g) of the lab, what would you estimate the percent error to be on the volume?

What percent error on the volume do you find if you use your answer to part A as the error on the volume?

In this case does the rule of thumb (part C) overestimate or underestimate the more precise error determination (part D) ?

a: Overestimate

b: Underestimate

+3

Answers (1)

Know the Answer?

Kaylie Chase · Answer 1

V = (8.2 ± 0.1) cm³, % = 1.2%

Explanation:

The volume of the sphere is

V = 4/3 π r³

The error in volume is

ΔV = 4 / 3 π r² Δr

DV = 4 π r² Δr

Let's calculate

V = 4/3 π (2.500 / 2) ³

V = 8,181 cm³

ΔV = 4 π 2,500²/4 0.005

ΔV = 0.098 cm³

V = (8.2 ± 0.1) cm³

The equation for the percentage error is

% = ΔV / V 100

% = 0.1 / 8.2 100

% = 1.2%