Question 255: Units , Dimensions and Error Analysis

Question

The radius of a sphere is measured to be (7.50 $\pm$ 0.85) cm. Suppose the percentage error in its volume is x.
The value of x, to the nearest x, is __________.

Accepted Answer

Given, radius of sphere = (7.50 $\pm$ 0.85) cm
$ 	herefore $ r = 7.50 and dr = 0.85
We know, volume of a sphere $v = {4 \over 3}\pi {r^3}$
taking log both sides, we get
$\ln v = \ln {{4\pi } \over 3} + 3\ln r$
Differentiating both sides,
${{dv} \over v} = 0 + 3{{dr} \over r}$
$ 	herefore $ Fractional error in volume ${{dv} \over v} = 3{{dr} \over r}$
$ 	herefore $ % error in volume,
${{dv} \over v} 	imes 100 = 3{{dr} \over r} 	imes 100$
$ = 3 	imes {{0.85} \over {7.50}} 	imes 100$
= 34%

Units , Dimensions and Error Analysis

The radius of a sphere is measured to be (7.50 \pm 0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest x, is __________.

Integer / Numeric Answer Type