Given :-

• ERROR in the measurement of radius of sphere = 2 %

To find :-

• error in the DETERMINATION of volume of sphere

Knowledge required :-

• Error in case of a measured quantity raised to a power

Generally ,

if Z = Aᵇ

then, Δ Z / Z = b ( Δ A / A )

Hence, the rule is STATED as : The relative error in a physical quantity raised to a power b is b times the relative error in the individual quantity.

Solution :-

As given

percentage error in radius , r = 2 %

so,

↦   ( Δ r / r )  = 2 % = 0.02 ....eqn (1)

Now,

Volume of sphere (V) = 4/3 π r³

( In this formula 4/3 and π are CONSTANT . hence, Error in the volume of sphere only depends on r³ )

So, using the rule for error in case of quantity raised to any power

↦ ( Δ V / V ) = 3 ( Δ r / r )

↦ ( Δ V / V ) = 3 ( 0.02 )

↦ ( Δ V / V ) = 0.06

↦ ( Δ V / V ) = 0.06

Hence,

Relative error in the measurement of volume of sphere is 0.06

Converting it to percentage error

↦ ( Δ V / V ) × 100 % = 0.06 × 100 %

↦ percentage error in the volume

of sphere = 6 %

Hence,

The percentage error in the determination of volume of sphere is 6 % .