The density of a sphere is measured by measuring its mass and diameter. If, it is known that the maximum percentage errors in the measurement are $$2\% $$ and $$3\% ,$$ then find the maximum percentage error in the measurement of density?
A.
$$15\% $$
B.
$$18\% $$
C.
$$9\% $$
D.
$$11\% $$
Answer :
$$11\% $$
Solution :
Let $$m$$ and $$d$$ be the mass and diameter of the sphere, then the density $$\rho $$ of the sphere is given by
$$\eqalign{
& \rho = \frac{{{\text{mass}}}}{{{\text{volume}}}} = \frac{m}{{\frac{4}{3}\pi {{\left( {\frac{d}{2}} \right)}^3}}} = \frac{{6m}}{{\pi {d^3}}} \cr
& {\text{or}}\,\,{\left( {\frac{{d\rho }}{\rho } \times 100} \right)_{\max }} = \left| {\frac{{dm}}{m} \times 100} \right| + \left| {\frac{{3d\left( d \right)}}{d} \times 100} \right| \cr
& = 2 + 3\left( 3 \right) \cr
& = 11\% \cr} $$
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