Question 227: 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