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Q.
The radius of a sphere is measured to be $(7.50 \pm 0.85)\,cm$. Suppose the percentage error
in its volume is $x$. The value of $x$, to the nearest $x$, is _______.
JEE MainJEE Main 2021Physical World, Units and Measurements
Solution:
$\because v =\frac{4}{3} \pi r ^{3}$
taking log then differentiate
$\frac{ dV }{ V }=3 \frac{ dr }{ r }$
$=\frac{3 \times 0.85}{7.5} \times 100 \%=34 \%$