Question

If the maximum error in the mea-
surement of the radius of a sphere!
is 2% then what is the maximum
error in the calculation of volume?​

Answer 1

Answer:

do you have the dimensions of the sphere

Answer 2

The maximum  error in the volume is 6%

Explanation:

Given that,

If the maximum error in the measurement of the radius of a sphere is 2% .

Let the radius of the sphere is r.

Let the volume of sphere is

$V = \dfrac{4}{3}\pi\times r^3$

We need to calculate the maximum  error in the radius

Using formula of error

$\dfrac{\Delta r}{r}=2\%$

$\dfrac{\Delta r}{r}=0.02$

We need to calculate the maximum  error in the volume

Using formula of error

$\dfrac{\Delta V}{V}=\dfrac{\dfrac{4}{3}\pi\times3r^2\times\Delta r}{\dfrac{4}{3}\times\pi r^2}$

$\dfrac{\Delta V}{V}=3\times\Delta r$

$\dfrac{\Delta V}{V}=3\times0.02$

$\dfrac{\Delta V}{V}=0.06$

$\dfrac{\Delta V}{V}=6\%$

Hence, The maximum  error in the volume is 6%

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