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

$9\%$

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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Single Correct Medium Published on 18th 08, 2020
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