First slide

Errors

Question

If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

Moderate

A

8%
B

2%
C

4%
D

6%

Solution

$V = \frac{4}{3} π R^{3}; \ln V = \ln (\frac{4}{3} π) + \ln R^{3}$
Differentiating, $\frac{d V}{V} = 3 \frac{d R}{R}$
Error in the determination of the volume
= 3 × 2% = 6%

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