Physics

If the error in measuring the radius of the sphere is $$2\%$$ and that in measuring its mass is $$3\%$$, then the error in measuring the density of material of the sphere is


ANSWER

$$9\%$$


SOLUTION
As density = $$\displaystyle \frac {mass}{volume}$$
$$\displaystyle \therefore\, \rho= \frac {M}{\frac {4}{3} \pi R^3}= \frac {3}{4} \frac {M}{4 \pi R^3}$$
$$\therefore$$ The percentage error in density is
$$\displaystyle \frac {\Delta \rho}{\rho} \times 100$$% $$\displaystyle = (\frac {\Delta M}{M} + 3 \frac {\Delta R}{R})\times 100$$%
=$$3$$% $$+ 3(2$$%$$) = 3$$% $$+ 6$$% $$= 9$$%
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