Question

If the error in the measurement of surface area of sphere is 4%. Then the error in the measurement of volume of sphere is
02%
o 4%
06%
08%

Answer 1

Given info : the error in the measurement of surface area of sphere is 4%.

To find : The error in the measurement of volume of sphere is ...

solution : surface area of sphere, A = 4πr² ....(1)

and volume of sphere , V = 4/3πr³ ...(3)

from equations (1) and (2) we get,

V = 4/3 π(A/4π)³/²

so, it is clear that, V ∝ A³/² ⇒ V² ∝ A³

differentiating both sides we get,

2V dV = 3A² dA

⇒2VdV/V² = 3A²/V² dA

⇒2dV/V = 3 A²/A³ dA

⇒2dV/V = 3dA/A

⇒2 dV/V × 100 = 3 dA/A × 100

⇒2 × % error in Volume = 3 × % error in area

given, % error in area = 4 %

so, 2 × % error in volume = 3 × 4 %

⇒% error in volume = 6 %

Therefore the error in the measurement of volume of sphere is 6 %