Question

if the error in the measurement in the eadius is 2 percentage then the error in the determination of volume of the sphere​

Answer 1

Answer:

• Percentage error in volume of sphere = 6 $\%$

Explanation:

Given,

• Error in the measurement of radius (R) is 2 $\%$

To find,

• Percentage error in the determination of volume (V) of sphere= ?

Solution,

• Given that percentage error in the measurement of radius is 2 $\%$ therefore, relative error will be 2/100 ; that is $\rm{\Delta R/R=2/100}$

Since, volume of sphere is given by

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

so,

Since, 4/3 and π are constant so, they will not have any effect in the error determination of volume hence, error in volume is dependent on r³ only.

• Using rule

[ The relative error in a quantity raised to a power k is given by k times the relative error in the individual quantity ]

therefore,

$\to \rm{\dfrac{\Delta V}{V}=3\times\bigg(\dfrac{\Delta R}{R}\bigg)}$

$\to \rm{\dfrac{\Delta V}{V}=3\times\bigg(\dfrac{2}{100}\bigg)}$

$\to \rm{\dfrac{\Delta V}{V}=\dfrac{6}{100}}$

so, We have relative error in the volume in sphere

Hence, Percentage error in the volume would be

$\to \rm{\dfrac{\Delta V}{V}\times 100\%=\dfrac{6}{100}\times 100\%}$

$\to \rm{\dfrac{\Delta V}{V}\times 100\%=6\;\%}$

Hence,

$\orange{\bigstar}\boxed{\boxed{\large{\rm{Percentage\:error\:in\:volume=6\;\%}}}}$