Question

Radius of spherical ball is 4.32 +- 0.01 cm. Calculate percentage error to estimate measurement of volume of sphere.

Answer 1

Explanation:

ANSWER

The volume is given as

V=(4/3)πr3

so, V= (4/3)×3.14×5.33

thus,

V = 623.30cm3

Now, the error equation

ΔV/V=3(Δr/r)

ΔV=V×3(Δr/r)

thus,

ΔV=623.30×3×(0.1/5.3)

ΔV=35.28cm3

thus,

ΔV/V×100=(35.28/623.30)×100

so,

ΔV/V×100=5.66%

Answer 2

Answer:

The percentage error to estimate the measurement of the volume of the sphere is 0.694%.                  

Explanation:

The percentage error in volume is calculated as,

$\mathrm{\frac{\Delta V}{V}=3\frac{\Delta R}{R}\times 100}$              (1)

Where,

ΔV=Change in the volume of the spherical ball

V=absolute volume of the spherical ball

ΔR=Change radius of the spherical ball

R=absolute radius of the spherical ball

From the question we have,

The radius of the spherical ball=4.32±0.01              (2)

Equation (2) is of the form:- R±ΔR                          (3)

Now by inserting all the required values in equation (1) we get;

$\mathrm{\frac{\Delta V}{V}=3\times \frac{0.01}{4.32}\times 100}$

$\mathrm{\frac{\Delta V}{V}=0.00694\times 100}$  

$\mathrm{\frac{\Delta V}{V}=0.694\%}$             (4)

So, the percentage error to estimate the measurement of the volume of the sphere is 0.694%.                  

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