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Q.
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
Physical World, Units and Measurements
Solution:
Volume of a sphere $=\frac{4}{3}\pi$ (radius)$^{3}$
or $V=\frac{4}{3}\pi R^{3}$
Taking logarithm on both sides, we have
log $V=$ log $\frac{4}{3}\pi+3$ log $R$
Differentiating, we get $\frac{\Delta V}{V}=0+\frac{3\Delta R}{R}$
Accordingly, $\frac{\Delta R}{R}=2\%$
Thus, $\frac{\Delta V}{V}=3\times2\%=6\%$